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Solution: Give the percent yield when 28.16 g of CO2 (molar mass = 44.01 g/mol) are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.2 C8H18 + 25 O2 → 16 CO2 + 18 H2OA) 20.00% B) 25.00% C) 50.00% D) 12.50%

Problem

Give the percent yield when 28.16 g of CO2 (molar mass = 44.01 g/mol) are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.

2 C8H18 + 25 O→ 16 CO2 + 18 H2O

A) 20.00% 

B) 25.00% 

C) 50.00% 

D) 12.50%

Solution

Use the concept of limiting reactant to determine the theoretical yield of CO2 when 4 moles of C8H8 and 4 moles of O2 reacts.

The given reaction is already balanced, all we need is the molar mass of CO2:

CO2        1 C × 12.01 g/mol C = 12.01 g/mol
               2 O × 16 g/mol O = 32 g/mol
                       Sum 
44.01 g/mol

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