How do i find the percent yield?

Question

Gaseous butane (CH₃(CH₂)₂CH₃) will react with gaseous oxygen (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O). Suppose 5.23 g of butane is mixed with 33.g of oxygen. Calculate the maximum mass of water that could be produced by the reaction. Round 3 sig figs.

Answer 1

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O ... balanced equation for combustion of butane

Since we are given the amounts of BOTH reactants, the first thing we must do is determine which reactant is limiting. One easy way to do this is to simply divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less will represent the limiting reactant:

For C₄H₁₀: 5.23 g x 1 mole / 58.12 g = 0.08999 moles (÷2->0.045)

For O₂: 33. g x 1 mole / 32 g = 1.031 moles (÷13->0.079)

Since 0.045 is less than 0.079, C₄H₁₀ is the limiting reactant and we will now use MOLES C₄H₁₀ to calculate moles of H₂O that can be produced by this reaction.

Use the stoichiometry of the balanced equation along with dimensional analysis to find the maximum mass of water produced:

0.08999 moles C₄H₁₀ x 10 mols H₂O / 2 mols C₄H₁₀ x 18.02 g H₂O / mol H2O = 8.11 g H₂O

This is the theoretical yield of water.

Percent yield = actual yield / theoretical yield (x100%)

The problem does not give the actual yield, so percent yield cannot be calculated.

Answer 2

Write and balance the chemical reaction equation:

2CH₃(CH₂)₂CH₃ + 13O₂ → 8CO₂ + 10H₂O

Calculate molar masses:

CH₃(CH₂)₂CH₃

C: 12.011 x 4 = 48.044

H: 1.008 x 10 = 10.080

Total: 58.124 g/mol

O₂

O: 15.999 x 2 = 31.998 g/mol

H₂O:

H: 1.008 x 2 = 2.016

O: 15.999 x 1 = 15.999

Total: 18.015 g/mol

Calculate the number of moles of the reactants:

CH₃(CH₂)₂CH₃:

(5.23 g)/(58.124 g/mol) = 0.08998 moles

O₂:

(33 g)/(31.998 g/mol) = 1.0313 moles

Calculate the limiting reactant:

It takes 2 moles of CH₃(CH₂)₂CH₃ for every 13 moles of O₂ so 0.08998 moles of CH₃(CH₂)₂CH₃ would require 0.585 moles of O₂ (divide 0.08998 by 2 and multiply the result by 13) so we have excess O₂ meaning CH₃(CH₂)₂CH₃ is the limiting reactant.

Calculate the number of moles of H₂O produced:

For every 2 moles of CH₃(CH₂)₂CH₃ there are 10 moles of H₂O produced. Therefore 0.08998 moles of CH₃(CH₂)₂CH₃ would yield 0.4499 moles of H₂O (divide 0.08998 by 2 and multiply the result by 10)

Calculate the mass of the H₂O produced:

(0.4499 mol)(18.015 g/mol) = 8.10495 g which is 8.10 g when rounded to 3 sig figs.

Percent Yield:

If you are looking for the % yield you would need to compare the actual amount produced (you didn't specify what that was) to the max possible (8.10495 g):

(actual # of grams produced)/(8.10495 g) x 100 = % yield