The question is incomplete; the complete question is;
Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 2.80g of carbon dioxide is produced from the reaction of 3.4g of octane and 5.9g of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.
Answer:
55%
Explanation:
The balanced equation is:
C8H18(g) + 25/2O2(g) ----------> 8CO2(g) + 9H2O(g)
The limiting reactant IS first determined. This is the reactant that yields the least amount of product.
3.4 g octane / 114 g/mol = 0.030 mol octane
From the reaction equation;
1 mole of octane yields 8 moles of CO2
0.030 mol octane yields 0.030 * 8/1 = 0.24 moles of CO2
5.9 g O2 / 32 g/mol = 0.18 mol O2
25/2 moles of O2 yields 8 moles of CO2
0.18 moles of O2 yields 8 * 0.18/25/2 = 0.1152 moles of CO2
Oxygen is the limiting reactant.
Number of moles of carbon dioxide produced (actual yield) = 2.8 g/44g/mol = 0.0636 moles of CO2
% yield = actual yield/ theoretical yield * 100
% yield = 0.0636/0.1152 * 100= 55%
