Answer:
Molecular formula = KSO4.
Empirical formula = K2S2O4.
Explanation:
What is given?
Mass of K (potassium) = 0.2361 g.
Mass of S (sulfur) = 0.1936 g.
Mass of O (oxygen) = 0.8162 g - 0.2361 g - 0.1936 g = 0.3865 g.
Molar mass of K = 39.1 g/mol.
Molar mass of S = 32 g/mol.
Molar mass of O = 16 g/mol.
Weight of compound = 270 g/mol.
Step-by-step solution:
Remember that empirical formulas show the simplest whole-number ratio of atoms in a compound, and molecular formulas show the number of each type of atom in a molecule. So first, let's see what would be the molecular formula:
Let's find the number of moles of each element using its molar mass:
[tex]\begin{gathered} 0.2361\text{ g K}\cdot\frac{1\text{ mol K}}{39.1\text{ g K}}=0.006038\text{ moles K,} \\ 0.1936\text{ g S}\cdot\frac{1\text{ mol S}}{32\text{g S}}=0.006050\text{ moles S,} \\ 0.3865\text{ g O}\cdot\frac{1\text{ mol O}}{16\text{ g O}}=0.02416\text{ moles O.} \end{gathered}[/tex]The next step is to divide each number of moles obtained by the least number of moles, in this case, the least number is 0.006038:
[tex]\begin{gathered} 0.006038\text{ moles K/0.006038=1 mol K,} \\ 0.006050\text{ moles S/0.006038 =}1.001\text{ moles S}\approx1\text{ mol S,} \\ 0.02416\text{ moles O/0.006038=4.001 moles O}\approx\text{4 moles O.} \end{gathered}[/tex]Our molecular formula would contain 1 mol of K, 1 mol of S, and 4 moles of O, so it would be KSO4.
But remember that we need to find the empirical formula too, based on the molecular weight. In the statement, they're telling us that the molecular weight is 270 g/mol, but the molecular weight of KSO4 is 135 g/mol (you can calculate it using the given molar masses and by doing an algebraic sum). So, to obtain the empirical formula with 270 g/mol as molecular weight, we have to multiply by two each number of moles of the compound (135 g/mol * 2 = 270 g/mol):
[tex]\begin{gathered} 1\text{ mol K}\cdot2=2\text{ moles K,} \\ 1\text{ mol S}\cdot2=2\text{ moles S,} \\ 4\text{ moles O }\cdot2=8\text{ moles O.} \end{gathered}[/tex]So, you can realize that K2S2O4 would be the empirical formula because its molecular weight would be 270 g/mol.
Molecular formula = KSO4.
Empirical formula = K2S2O4.
How many moles of chromium metal, Cr, are in a 260 gram piece of chromium?
In order to answer this question we will use the molar mass of Chromium, which is 52 g/mol, that means that in every 1 mol of Cr, we will have 52 grams of it:
52 g = 1 mol
260 g = x moles
52x = 260
x = 5 moles of Chromium
How to find oxidation number of ReO4^-?
The oxidation number of O is usually -2 and in this case it is.
Our ion is ReO₄⁻. It has a total charge of -1. The charge of the four atoms of O is -8. With this information we can write an equation. X will represent the oxidation state of Re.
Total charge = Oxidation state of Re + 4 * Oxidation state of O
- 1 = x + 4 * (-2)
- 1 = x - 8
-1 + 8 = x
x = +7
So the oxidation state of Re is +7.
Which of the following set of quantum numbers are NOT possible? (You can pick more than one)
a. 1, 0, 0, +1/2
b. 4, 0, 0, +1/2
c. 3, 3, -3, -1/2
d. 2, 1, 1, -1/2
e. 2, 1, 2, +1/2
The set of quantum numbers (3, 3, -3, -1/2), and (2, 1, 2, +1/2) are not possible. Therefore, options c and e are correct.
What are the quantum numbers?The set of numbers that can describe the position and energy of the electron. We have four quantum numbers principal, azimuthal, magnetic, and spin quantum numbers.
Principal quantum numbers (n) tell about the principal electron shell of the atom and the most probable distance between the electrons and the nucleus. The azimuthal quantum number (l) tells about the shape of an orbital in which an electron is present. It has values equal to l = n - 1.
The magnetic quantum number tells about the total number of orbitals in a subshell and the orientation of these orbitals. The value of m lies in between -l to +l. The value of the spin quantum number describes the direction of spin of an electron and it has only two values +½ and -½.
In option c, the principle (n) and azimuthal (l) quantum number has the same value but when n = 3 the possible values of l are 2, 1, 0. In option d, the magnetic quantum number has the value of 2 while l has 1 which is not possible because when l= 1 the values of m can be +1, 0, and -1.
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If39.50 ml of a vegetable juice contains 45% of the recommendeddaily allowance of vitamin C (equal to 60 mg). How many milliliters ofthe vegetable juice will provide 100% of the recommended dailyallowance?
87.77 ml will provide the recommended daily allowance of vitamin C.
- To calculate the milliliters of vegetable juice that will provide 100% of the recommended daily allowance of vitamin C, we can use a mathematical Rule of Three:
[tex]\frac{100.39.50}{45}=87.77[/tex]So, 87.77 ml will provide the recommended daily allowance of vitamin C.
Starting with a gas of N2 in a balloon of temperature 148.5°C and volume 241.8mL, what is its final volume if you cool it to -96.4°C?
Answer
101.3 mL
Explanation
Given:
The initial temperature, T₁ = 148.5 °C = (148.5°C + 273) = 421.5 K
The initial volume, V₁ = 241.8 mL
Final temperature, T₂ = -96.4 °C = (-96.4°C + 273) = 176.6 K
What to find:
The final volume of the gas.
Step-by-step solution:
The final volume, V₂ of the gas can be calculated using Charle's law formula.
[tex]\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \\ \Rightarrow V_2=\frac{V_1\times T_2}{T_1}=\frac{241.8mL\times176.6K}{421.5K}=\frac{42701.88\text{ }mL}{421.5} \\ \\ V_2=101.3\text{ }mL \end{gathered}[/tex]The final volume of the gas is 101.3 mL
How many moles are in 53.99 mg chromium?
I’m not sure and I’m kind of confused can anyone help?
We will reconstruct the model in the following manner :
From the above diagram we can see that :
• number of Carbon atom = 3
• number of hydrogen atom = 8
• rewrite this in an alphabetical order, you get :
[tex]\begin{gathered} C_3H_8\text{ } \\ \Rightarrow Propane\text{ } \end{gathered}[/tex]the molecule has a chemical formula = C3H8What is the density of hydrogen sulfide (H2S) at 0.2 atm and 311 K?Answer in units of g/L
Answer
Density = 0.267 g/L
Explanation
Given:
Pressure of H2S = 0.2 atm
Temperature = 311 K
We know:
The molar mass of H2S = 34,1 g/mol
R constant = 0.08206 L.atm/K.mol
Solution:
From the ideal gas law:
PV = nRT
We know that:
density = m/V
n = m/M
Therefore we can use the following equation to solve for density of H2S
[tex]\begin{gathered} density\text{ = }\frac{PM}{RT} \\ density\text{ = }\frac{(0.2\text{ atm x 34,1 g/mol\rparen}}{(0.08206\text{ }L.atm/K.mol\text{ x 311 K\rparen}} \\ \\ density\text{ = 0.267 g/L} \end{gathered}[/tex]
1. 60.0 mL of 0.322 M lithium chloride, LICI (aq) are combined with 20.0 mL of 0.530 M Tin (II) nitrate,
Sn(NO₂)2 (aq), 0.632 g of precipitate are recovered.
a. Write a balanced chemical equation for the reaction.
b. Write a balanced net-ionic equation for the reaction.
c. Calculate the moles of precipitate that are actually produced in the reaction.
d. Calculate the moles of precipitate that should be produced if the reaction went to completion.
e. Calculate the percent yield of the reaction.
Answer:
60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate.
How many grams of lead (II) iodide will precipitate? (you must write your own reaction)
Explanation:
A reaction experimentally yields 15.68 g of a product. What is the percent yield if the theoretical yield is 18.81 g?
Answer
The percent yield = 83.36%
Explanation
Given:
Experimental yield = actual yield = 15.68 g
Theoretical yield = 18.81 g
What to find:
The percent yield for the reaction.
Step-by-step solution:
The percent yield for the reaction can be calculated using the formula below:
[tex]\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{15.68\text{ }g}{18.81\text{ }g}\times100\% \\ \\ Percent\text{ }yield=83.36\% \end{gathered}[/tex]Hence, the percent yield for the reaction is 83.36%
54000 mL isO 54 m3O 5400 cm3O 0.054m3O 540 m3
mL and cm³ have a 1 to 1 conversion, so we have:
[tex]54000mL=54000cm^{3}[/tex]But we don't have this option, so we will need to find another.
The other are in m³, so we can use the conversion from cm to m to get this:
[tex]\begin{gathered} 1m=100cm \\ (1m)^{3}=(100cm)^{3} \\ 1m^{3}=1000000cm^{3} \\ 1cm^{3}=\frac{1}{1000000}m^{3} \end{gathered}[/tex]So, we can apply this to what we have:
[tex]54000cm^3=54000\cdot\frac{1}{1000000}m^3=0.054m^{3}[/tex]We have an option with 0.054m³, so the correc alternative is 0.054 m³.
8. Based on the Law of Conservation of Matter: At the start of the reaction 20g of
one material and some amount of another material were reacted and produced
30g of solid and 70g of a gas. What is the other amount of reactant used?
a. 20
b. 60g
c. 80g
d. 100g
Answer:
80g
Explanation:
The law of conservation of matter states that matter cannot be created nor destroyed. Basically, whatever mass you have at the beginning of a reaction, should be the same as at the end of the reaction on the product side.
Since here it lets you know that a total of 100g (30g + 70g) were produced on the product side, that means that we started off with 100g in our reactant side.
We have given that one material is 20g on our reactant side but we need the mass of the other. To find the mass of the other material, simply subtract 20g from the total mass created on the product side.
100g - 20g = 80g
The 80g would be the missing amount from the reactant side that isn't stated.
Can someone help me to answer this?As you reflect on how to interpret a balanced chemical reaction.Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 2002, Philippines has banned imports of ammonium nitrate that used in bombs that killed 12 people in Mindanao area.The explosion resulted from this reaction:2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g)Construct a table showing how to interpret the information in the equation in terms of:1. individual molecules and ions.2. moles of reactants and products.3. grams of reactants and products given 2 mol of ammonium nitrate.4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.
(1)
The molecule NH4NO3 is ammonium nitrate, which makes a redox reaction.
It produces N2, O2, and H2O, which are nitrogen, oxygen, and water.
The ions are 2NH4+ and 2NO3-.
2 molecules of N2, 4 molecules of H2O, and 1 molecule of O2.
(2)
There are 2 moles of NH4NO3, 2 moles of N2, 4 moles of H2O, and 1 mole of O2.
(3)
The molar mass of NH4NO3 is 80.043 grams per mole, but there are 2 moles of it, so there are 160.09 grams of NH4NO3.
There are 56.03 grams of N2 because there are 2 moles of it. (1 mole N2 = 28.0134 g/mol).
There are 72.06 grams of H2O because there are 4 moles of it (1 mole H2O = 18.02 g/mol).
There are 31.998 grams of O2 because there's just 1 mole of it.
(4)
The formula units of NH4NO3 is 1.204x10^24, which is equivalent to 2 moles.
Nitrogen has the same formula units because there are 2 moles of it, so it's 1.204x10^24.
Water has 2.409x10^24 because there are moles of it.
Oxygen has 6.022x10^23 because there's just 1 mole, inis Avogadro's Number.
The quantatum mechanical model of an atom uses atomic orbitals to describe what
If you wanted to dilute the 3M NaOH solution to 500mL of 1M NaOH solution, how much L of the 3M NaOH solution would you need?
0.167L
Explanation:In order to know how much L of the 3M NaOH solution would you need, we will simply set up an equal proportion solution expressed as;
[tex]C_1V_1=C_2V_2[/tex]C1 and C2 are the concentration of the solutions
V1 and V2 are the volumes of the solutions
Given the following parameters;
C1 = 1M
V1 = 500mL
C2 = 3M
V2 = ?
Substitute the given parameters into the formula above to get the required litre of solution needed.
[tex]\begin{gathered} 1\times500=3\times V_2_{} \\ 500=3V_2 \\ \text{Swap} \\ 3V_2=500 \end{gathered}[/tex]Divide both sides by 3:
[tex]\begin{gathered} \frac{3V_2}{3}=\frac{500}{3}_{} \\ V_2=166.67mL \end{gathered}[/tex]Converting to litres
Since 1mL = 0.001L
166.67mL = x
Cross multiply
[tex]\begin{gathered} 1\times x=166.67\times0.001 \\ x=0.167L \end{gathered}[/tex]Hence the amount of L of the 3M NaOH solution would you need is 0.167L
Co-60 is used medically for radiation therapy as implants and as an external source of radiation exposure. The half-life of
Co-60 is 5, 272 years. How much of a 2.000 mg sample will remain after 21, 088 years? You must show your work to receive
credit.
. Show the equation needed
b. Show a picture of you solving for the unknown.
c. Show the final answer
Co-60 is used medically for radiation therapy as implants and as an external source of radiation exposure the half-life of Co-60 is 5, 272 years 2.000 mg sample will remain after 21, 088 years is 0.0625 gm left
Radiation therapy is the cancer treatment that uses high doses of radiation to kill cancer cell and shrink tumor
Here given data is Co-60 is 5, 272 years and we have to find the number of half lives in 21, 088 years = ?
21, 088/5, 272 = 4 half lives
(1/2)⁵ = 1/32 nd of the original will be left
1/32×2.000mg = 0.0625 gm left
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A 1.6700 g sample of pure compound was analyzed and found to contain 0.1870 ghydrogen, 0.8899 g carbon with the remainder being oxygen. What is the molecularformula of the unknown compound if its molar mass is 90.14 g/mol.?
Answer: C4H10O2
Explanation:
At first find the percentage composition, in 1.67 g compound what percentage of Carbon, Hydrogen and Oxygen are there. Then from those percentages through unitary method find out how much Carbon, Hydrogen and Oxygen would be there in 90.1 g of compound. From those amounts finds their mole number, as in divide the obtained amount by their gram equivalent mass. The numbers you obtain are the corresponding number of Carbon Hydrogen and Oxygen in one molecule of the compound. Hope this helps
Substance Density (grams/cm3)Chloroform - 1.5Ebony wood - 1.2Mahogany wood - 0.85Oil - 0.9Water - 1.025.Since volume = mass/density, a 1,700 gram beam of mahogany wood has a volume of...Volume = Mass / DensitySelect one:a. 500 cm3b. 1,445 cm3c. 1,785 cm3d. 2,000 cm3
As the question gave us the formula in which we have to use to calculate the volume of this type of wood:
V = m/d
We have:
m = 1700 grams
d = 0.85
Now we add these values into the formula:
V = 1700/0.85
V = 2000 cm3, letter D
What functional group is found in amino acids?A) aminesB) alkanesB) cyclic hydrocarbon ringsC) alcohols
Answer:
[tex]A)\text{ Amines}[/tex]Explanation:
Here, we want to get the functional group is found in amino acids
In the amino acids, we can get amines, carboxylic acid groups, and the carbon chain
Looking at the options, we can see that the amine group is the right option here as the other two are not available
What a balanced chemical equation for the single displacement reaction you observed in Experiment 3. Include physical states.
a.
__K (s) + __Cl2 (g) —> __ KCl (aq)
Balance, and Type of reaction?
[tex]2K_(s) + Cl_2_(g)[/tex] ⇒ [tex]2KCl_(aq)[/tex]
Type of Reaction: Synthesis
What is a synthesis reaction?
When 2 or more substances react to form a single compound.
Which two elements would have the same number of energy levels
The elements with the same number of energy levels are the ones that belong to the same period of the periodic table.
Copper and Zinc both belong to the 4th period of the periodic table.
Nickel belongs to the 4th period and palladium to the 5th period.
Lithium belongs to the 2nd period and Magnesium to the 3rd period.
It means that the correct answer is Copper and Zinc, they belong to the same period,
A sample of glass that has a mass of 9.3 g gives off 87 J of heat. If the temperature of the sample changes by 12.7°C during this change, what is the specific heat of the glass? NEED ASAP
Answer
The specific heat of the glass = 0.7366 J/g°C
Explanation
Given:
Mass of the glass sample, m = 9.3 g
Quantity of heat given off, Q = 87 J
Change in temperature, ΔT = 12.7 °C
What to find:
The specific heat, c of the glass.
Step-by-step solution:
The specific heat, c of the glass can be calculated using the given formula below.
[tex]\begin{gathered} Q=mc\Delta T \\ \\ \Rightarrow c=\frac{Q}{m\Delta T} \end{gathered}[/tex]Putting the values of the given parameters into the formula, we have;
[tex]c=\frac{87J}{9.3g\times12.7°C}=\frac{87\text{ }J}{118.11\text{ }g°C}=0.7366\text{ }J\text{/}g°C[/tex]Therefore, the specific heat of the glass is 0.7366 J/g°C
A student wants to produce a 1.8 M solution and has 0.9 moles of solute available. What is the maximum volume (in mL) of solution that can be produced?Given:Find:Equation used:Answer:
Answer:
[tex]500\text{ mL}[/tex]Explanation:
Here, we want to get the maximum volume of solution that can be produced
Given:
Molarity = 1.8 M
Number of Moles = 0.9 moles
Find:
Volume
Equation Used:
Number of moles = molarity * volume
Answer:
[tex]\begin{gathered} 0.9\text{ = 1.8 }\times\text{ V} \\ V\text{ = }\frac{0.9}{1.8} \\ V=0.5dm^3 \end{gathered}[/tex]To convert this to mL, we multiply the volume by 1000 since 1 L = 1000 mL
Thus, we have it that:
[tex]0.5\text{ }\times1000\text{ = 500 mL}[/tex]If a student did an experiment were they mixed 25 mL of HCl (0.5M) and 7.4g of sodium bicarbonate together How many moles of carbon dioxide would be generated at STP? Show your balanced equation aswell.
The first thing will be to finish the reaction that happens in the experiment to the reactional HCl and sodium bicarbonate (NaHCO3). The balanced reaction is the following:
[tex]NaH_{}CO_3+HCl\rightarrow NaCl+H_2O+CO_{2(g)}[/tex]Now, we must determine which is the limiting reactant. To do this we are going to convert all the data they give us to moles.
For HCl we are given the molarity, so the moles will be:
[tex]\begin{gathered} \text{Molarity}=\text{ }\frac{\text{Moles of solute}}{L\text{ of solution}} \\ \text{Moles of solute= Molarity}\times L\text{ of solution} \\ \text{Moles of solute=}0.5M\times(25mL\times\frac{1L}{1000mL}) \\ \text{Moles of solute=}0.0125\text{ mol HCl} \end{gathered}[/tex]Now the moles of sodium bicarbonate are found using its molar mass:
[tex]\begin{gathered} \text{Moles }NaHCO3=\text{Given g }NaHCO3\times\frac{1molNaHCO3}{Molar\text{ mass, g}NaHCO3} \\ \text{Moles }NaHCO3=7.4g\times\frac{1moNaHCO3}{84.007gNaHCO3} \\ \text{Moles }NaHCO3=0.0880\text{mol }NaHCO3 \end{gathered}[/tex]By stoichiometry, we have that one mole of NaHCO3 reacts with one mole of HCl. We have more moles of NaHCO3 than HCl. Therefore, HCl will be the limiting reactant.
So the reaction will occur according to the number of moles of HCl. Now the ratio between HCl and CO2 gas formed is 1 to 1. For one mole of HCl that reacts, 1 mole of CO2 will be formed.
so, the number
Sugar forms when carbon, oxygen, and hydrogen combine in a specific ratio. From what you know about elements and the periodic table, what is true about the bonding in sugar?
Answer
B. The bonds in the compound are covalent
Explanation
The carbon, oxygen, and hydrogen that combine in a specific ratio to form the sugar are nonmetals. From what was learned about elements and the periodic table, nonmetals form covalent bonds among themselves to form compounds.
So what is true about the bonding in sugar is:
B. The bonds in the compound are covalent.
A solution with a total volume of 1000.0 mL contains 37.1 g Mg(NO3)2. If you remove 20.0 mL of this solution and then dilute this 20.0 mL sample with water until the new volume equals 500.0 mL, what is the concentration of Mg+2 ion in the 500.0 mL of solution? What is the concentration of nitrate ion?
1. The concentation of the magnesium ion, Mg²⁺ in the solution is 0.01 M
2. The concentation of the nitrate ion, NO₃⁻ in the solution is 0.02 M
We'll begin by obtaining the concentration of the stock solution. This can be obtained as follow:
Mass of Mg(NO₃)₂ = 37.1 gMolar mass of Mg(NO₃)₂ = 148 g/moleMole of Mg(NO₃)₂ = 37.1 / 148 = 0.25 moleVolume = 1000 mL = 1000 / 1000 = 1 LConcentration =?Concentration = mole / volume
Concentration = 0.25 / 1
Concentration = 0.25 M
Next, we shall determine the concentration of the diluted solution
Volume of stock solution (V₁) = 20 mLConcentration of stock solution (C₁) = 0.25 MVolume of diluted solution (V₂) = 500 mL Concentration of diluted solution (C₂) =?C₁V₁ = C₂V₂
0.25 × 20 = M₂ × 500
5 = M₂ × 500
Divide both side by 500
C₂ = 5 / 500
C₂ = 0.01 M
1. How to determine the concentration of magnesium ion, Mg²⁺
Mg(NO₃)₂(aq) <=> Mg²⁺(aq) + 2NO₃⁻(aq)
From the balanced equation above,
1 mole of Mg(NO₃)₂ contains 1 mole of Mg²⁺
Therefore,
0.01 M Mg(NO₃)₂ will also contains 0.01 M Mg²⁺
Thus, the concentration of Mg²⁺ is 0.01 M
2. How to determine the concentration of nitrate ion, NO₃⁻
Mg(NO₃)₂(aq) <=> Mg²⁺(aq) + 2NO₃⁻(aq)
From the balanced equation above,
1 mole of Mg(NO₃)₂ contains 2 mole of NO₃⁻
Therefore,
0.01 M Mg(NO₃)₂ will contain = 0.01 × 2 = 0.02 M NO₃⁻
Thus, the concentration of NO₃⁻ is 0.02 M
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Density of gasses lawA sample of gas with a density of 0.066 g/l at 32 C and a pressure of 0.95 atm. Find the density of the gas at STP
Answer:
[tex]0.078\text{ g/l}[/tex]Explanation:
Here, we want to calculate the density of the gas at STP
We use a modification of the general gas law as follows
Mathematically:
[tex]\frac{P_1}{D_1\times T_1}\text{ = }\frac{P_2}{D_2\times T_2}[/tex]where:
P1 is the initial pressure which is 0.95 atm
D1 is the initial density which is 0.066 g/l
T1 is the initial temperature in Kelvin (we add the temperature in Celsius with 273 K : 32 + 273 = 305 K)
P2 is the pressure at STP which is 1 atm
D2 is the density that we want to calculate
T2 is the temperature at STP which is 273 K
Substituting the values, we have it that:
[tex]\begin{gathered} \frac{0.95}{0.066\times305}\text{ = }\frac{1}{D_2\times273} \\ \\ D_2\text{ = }\frac{0.066\times305}{0.95\times273}\text{ = 0.078 g/l} \end{gathered}[/tex]How many molecules of ethane gas, C2H6 are in 15 grams of the compound?
3.01×10²³molecules.
Explanations:
The formula for the number of molecules of a compound given the number of moles is expressed as:
[tex]nu\text{mber of molecules=moles}\times6.02\times10^{23}[/tex]Get the moles of ethane gas using the formula:
[tex]\begin{gathered} \text{moles of ethane=}\frac{Mass\text{ of ethane}}{Molar\text{ mass of ethane}} \\ \text{Moles of ethane=}\frac{15}{2(12)+1(6)} \\ \text{Moles of ethane}=\frac{15}{30}\text{moles} \\ \text{Moles of ethane}=0.5\text{moles} \end{gathered}[/tex]Determine the required number of molecules of ethane
[tex]\begin{gathered} nu\text{mber of mol}ecules=0.5\times6.02\times10^{23} \\ nu\text{mber of mol}ecules=3.01\times10^{23} \end{gathered}[/tex]Hence the molecule of ethane gas that is in 15 grams of the compound is 3.01×10²³molecules.
Methane(CH4) gas and oxygen (O2) gas react to form carbon dioxide (CO2) gas and water vapor(H2O). Suppose you have 5.0 mol of CH4 and 1.0 mol of O2 in a reactor.
What would be the limiting reactant? Enter its chemical formula below
The limiting reactant, given that 5.0 moles of CH₄ and 1.0 mole of O₂ are in the reactor is O₂
How do I determine the limiting reactantWe'll begin by obtainig the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
The limiting reactant for the reaciont can be obtained as illustrated below:
From the balanced equation above,
1 mole of CH₄ reacted with 2 moles of O₂
Therefore,
5 moles of CH₄ will react with = 5 × 2 = 10 moles of O₂
From the above illustration, we can see that a higher amount of O₂ is needed to react completely with 5 moles of CH₄.
Thus, we can conclude that O₂ is the limiting reactant.
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