Ishwar S. answered • 05/30/19
University Professor - General and Organic Chemistry
2 H2 (g) + O2 (g) → 2 H2O (g)
Convert g of H2 and g of O2 to g of H2O, respectively. The reactant that produces the LOWER amount of H2O is the limiting reagent, and that is the mass of H2O produced from this reaction.
10.0 g H2 x (1 mol H2 / 2 g H2) x (2 mol H2O / 2 mol H2) x (18 g H2O / 1 mol H2O) = 90.0 g H2O
75.0 g O2 x (1 mol O2 / 32 g O2) x (2 mol H2O / 1 mol O2) x (18 g H2O / 1 mol H2O) = 84.4 g H2O
O2 is the limiting reagent, and 84.4 g of H2O is produced from this reaction.
H2 is therefore, the excess reactant. Let's calculate how much H2 reacts with 75.0 g of O2.
75.0 g O2 x (1 mol O2 / 32 g O2) x (2 mol H2 / 1 mol O2) x (2 g H2 / 1 mol H2) = 9.38 g H2
Out of 10.0 g H2 used, 9.38 g of H2 reacts. Mass of H2 gas remaining = 10.0 - 9.38 = 0.62 g H2 leftover
