Please use the below balanced equation to answer this question. 2H2(g) + O2(g) -> 2H2O(l) How many grams of water will be produced if I have 83.4 liters of oxygen gas at STP?
A) 48.65 grams H2O
B) 65.78 grams H2O
C) 85.71 grams H2O
D) 134.18 grams H2O
At STP (standard temperature and pressure conditions), 1 mol of any gas occupies 22.4 L This rule is applied to O₂ 22.4 L volume occupied by 1 mol Therefore 83.4 L occupied by - 1/ 22.4 x 83.4 = 3.72 mol stoichiometry of O₂ to H₂O is 1:2 then the number of moles of water produced - 3.72 mol x 2= 7.44 mol mass of water produced - 7.44 mol x 18.01 g/mol = 134.1 g correct answer is D
