Question

For the following reaction, 4.62 grams of oxygen gas are mixed with excess benzene (CH). The reaction yields 3.89 grams of carbon dioxide benzene (CH) (1) + oxygen (g) carbon dioxide (g) + water (g) What is the theoretical yield of carbon dioxide ? What is the percent yield of carbon dioxide?

Answer 1

Balanced equation for this reaction is ,

2C6H6 + 15O2 ------------> 12CO2 + 6H2O

15 moles of oxygen gives 12 moles of carbon dioxide

1 mole of oxygen will give 12/15 i.e. 0.8 moles of carbon dioxide.

Number of moles = mass/molar mass

Molar mass of O2 = 32 g/mol

Moles of O2 = 4.62/32

= 0.144 moles

Moles of CO2 formed = 0.144 × 0.8

= 0.115 moles CO2

Molar mass of CO2 = 44 g/mol

Mass of CO2 formed = 44 g/mol × 0.115 mol

= 5.06 grams

Theoretical yield = 5.06 g

Actual yield = 3.89 g

Percent yield = (actual yield/theoretical yield)×100

= (3.89/5.06)×100

= 76.88 %

Percent yield = 76.88 %