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The reaction of 6.80g of carbon with excess O2 yields 12.7 g of CO2. What is...

The reaction of 6.80g of carbon with excess O2 yields 12.7 g of CO2. What is the percent yield of this reaction?
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Answer #1

Molar mass of C = 12.01 g/mol

mass of C = 6.8 g

mol of C = (mass)/(molar mass)

= 6.8/12.01

= 0.5662 mol

we have the Balanced chemical equation as:

C + O2 —> CO2

From balanced chemical reaction, we see that

when 1 mol of C reacts, 1 mol of CO2 is formed

mol of CO2 formed = moles of C

= 0.5662 mol

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.5662*44.01

= 24.92 g

% yield = actual yield * 100 / theoretical yield

= 12.7*100/24.92

= 51.0 %

Answer: 51.0 %

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