Question

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

Answer 1

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