Study the lab scenario and balanced chemical reaction.
An excess amount of butane undergoes a combustion reaction to produce CO2 and H2O. 4.62 x 101 g of oxygen gas are consumed, at STP.
2C4H10 + 13O2 → 10H2O + 8CO2
If the actual yield is 93.7% of the theoretical yield, what is the actual yield, in g, of water?
1.67 x 101 g of H2O
1.88 x 101 g of H2O
2.68 x 100 g of H2O
3.57 x 101 g of H2O
Is it C?
46.2g O2 = 1.44 moles
so, 10/13 * 1.44 * 18 * 0.937 = 18.68 g O2
Looks like B to me