Calculate the mass of water produced when 1.47 g of butane reacts with excess oxygen.

Answer 1

2C4H10 + 13 O2 ==> 8CO2 + 10H2O
mols C4H10 = grams/molar mass = 1.47/58 = ?
mols H2O produced = ? from previous step x (10 mols H2O/2 mols C4H10) = ? from previous step x (10/2) = xx
Then convert to grams. grams = xx mols H2O x molar mass H2O
Post your work if you get stuck.

Answer 2

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