Question #55701

The gaseous hydrocarbon butane, C4H10, burns according to the following equation:
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)
How many grams of CO2 are produced from the complete reaction of 55.6 g of C4H10?
Express your answer with the appropriate units.
m(CO2) = 168 g (already answered)
Part D
If the reaction in part c produces 130 g of CO2, what is the percent yield of CO2 for the reaction?
Express your answer with the appropriate units.
m(O2) = ????? (this is what I need to find)

Expert's answer

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

m(O2) = M(O2)*n(O2) = M(O2)*13/8*n(CO2) = 32*13/8* n(CO2)

n(CO2) = m(CO2)/M(CO2) = 130(g)/44(g/mol)= 2.955 mol

m(O2) = 32*13/8*2.955 = 153.64 g

Answer : m(O2) = 153.64 g

m(O2) = M(O2)*n(O2) = M(O2)*13/8*n(CO2) = 32*13/8* n(CO2)

n(CO2) = m(CO2)/M(CO2) = 130(g)/44(g/mol)= 2.955 mol

m(O2) = 32*13/8*2.955 = 153.64 g

Answer : m(O2) = 153.64 g

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