2C4H10 + 13O2 ==> 8CO2 + 10H2O
moles C4H10 = grams/molar mass = 6.51/58 = 0.112
mols H2O produced = 0.112 mols C4H10 x (10 moles H2O/2 mols C4H10) =
0.112 x 5 = 0.56
g H2O = mols x molar mass = ?
Calculate the mass of water produced when 6.51 g of butane reacts with excess oxygen.
1 answer