The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water. 2 C4H10 + 13 O2 ------- 8 CO2 + 10 H2O (a) How many moles of water formed? (b) How many moles of butane burned? (c) How many grams of butane burned? (d) How much oxygen was used up in moles? In grams?

Question

The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water. 2 C4H10 + 13 O2 -------> 8 CO2 + 10 H2O (a) How many moles of water formed? (b) How many moles of butane burned? (c) How many grams of butane burned? (d) How much oxygen was used up in moles? In grams?

anonymous · Answer

(a) 
(2.46 g H2O) / (18.01532 g H2O/mol) = 0.137 mol H2O 
(b) 
(2.46 g H2O) / (18.01532 g H2O/mol) x (2 mol C4H10 / 10 mol H2O) = 
0.0273 mol C4H10 
(c) 
(0.0273 mol C4H10) x (58.1222 g C4H10/mol) = 
1.59 g C4H10 
(d) 
(2.46 g H2O) / (18.01532 g H2O/mol) x (13 mol O2 / 10 mol H2O) = 
0.178 mol O2 
(0.178 mol O2) x (31.99886 g O2/mol) = 5.70 g O2