Answer to Question #95566 in Chemistry for Bruno

Question

Answer the following questions by referring to the equation below:

2 C6H10 + 17 O2  12 CO2 + 10 H2O

If 35 grams of C6H10 and 45 grams of oxygen are reacted,
i) which one is the limiting reagent?
ii) how many grams of carbon dioxide will be formed?
iii) how much of the excess reagent is left over after the reaction is complete?
iv) if 35 grams of carbon dioxide are formed, calculate the percent yield of this reaction.

Answer 1

i) Number of moles of reagents must be estimated to determine the limiting reagent:

n = m/Mr,

where m - mass, Mr - molecular weight, n - number of moles.

From here:

n (C₆H₁₀) = 35 g / (82 g/mol × 2) = 0.2 mol

n (O₂) = 45 g / 544 = 0.08 mol

Answer: O₂ is a limiting reagent.

ii) As O₂ is a limiting reagent, 0.08 mol of CO₂ will be produced. From here:

m = n × Mr = 0.08 mol × 528 g/mol = 42.24 g.

Answer: 42.24 g of CO₂.

iii) Only 0.08 mol of C₆H₁₀ will be used in the reaction. As a result, 0.2 mol - 0.08 mol = 0.12 mol of C₆H₁₀ remain after the reaction is complete. From here:

m = n × Mr = 0.12 mol × 164 g/mol = 19.68 g

Answer: 19.68 g of C₆H₁₀.

iv) The theoretical yield of CO₂ is 42.24 g and actual yield is 35 g. The percent yield of the reaction is:

% yield = (actual yield / theoretical yield) × 100 % = (35 g / 42.24 g) × 100 % = 83 %.

Answer: 83 %.