1 C3H8 + 5 O2® 4 H2O + 3 CO2 32 grams of propane (C3H8) is burned in excess oxygen gas to produce how many grams of water?

Question

a
52.4
b
26.8
c
37.6
d
13.7

Answer 1

Since the problem states that here is excess oxygen, one need not determine the limiting reactant.  From the balanced equation, one notes that 1 mole C3H8 produces 4 moles H2O.  So, first we convert 32 grams of propane to moles and then find moles of H2O.  Then convert moles of H2O to grams of H2O

Moles of H2O produced = 32 g C3H8 x 1 mole/44 g x 4 moles H2O/mole C3H8 = 2.909 moles H2O

Grams H2O produced = 2.909 moles H2O x 18 g/mole = 52.36 g = 52 g H2O (to 2 significant figures)