Butane reacts with oxygen to produce carbon dioxide and water following the equation given below. C4H10(g) + 132O2(g) 4CO2(g) + 5H2O(l) If 174.0 kg of butane is mixed with 320.0 kg of O2, the volume of water formed in liters is . (Nearest integer) [Given : (a) Molar mass of C, H, O are 12, 1, 16 g mol-1 respectively, (b) Density of water = 1 g mL-1]
Answer
138
Solution
Molar mass of C4H10 = 58 g mol-1, of O2 = 32 g mol-1
Moles of C4H10 = 174.0 10358 = 3000 mol
Moles of O2 = 320.0 10332 = 10000 mol
O2 needed for 3000 mol butane = 3000 132 = 19500 mol > 10000 mol
O2 is the limiting reagent.
Moles of H2O = 10000 513/2 = 10000013 = 7692.3 mol
Mass of H2O = 7692.3 18 = 1.3846 105 g
Volume = 1.3846 105 mL = 138.46 L 138 L
Answer: 138