Question #55703

The gaseous hydrocarbon butane, C4H10, burns according to the following equation:
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)
How many grams of O2 are needed to react completely with 23.5 g of C4H10?
Express your answer with the appropriate units.
m(O2) = ?????
Thank you so much!!

Expert's answer

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

m(O2) = M(O2)*n(O2)

n(O2) = (13/2)*n(CO2) = (13/2)*(m(CO2)/M(CO2)) = (13/2)*(23.5/58) = 2.63 mol

m(O2) = M(O2)*n(O2) = 32*2.63 = 84.28 g

Answer : m(O2) = 84.28 g

m(O2) = M(O2)*n(O2)

n(O2) = (13/2)*n(CO2) = (13/2)*(m(CO2)/M(CO2)) = (13/2)*(23.5/58) = 2.63 mol

m(O2) = M(O2)*n(O2) = 32*2.63 = 84.28 g

Answer : m(O2) = 84.28 g

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