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

83.4 L O2 at STP
know that for every 22.4 L of gas at STP is 1 mole
83.4 L / 22.4 L = 3.72 mol O2

Then according to the reaction; for every 1 O2 you get 2 H2O

3.72 mol O2 * (2 H2O / 1 O2) = 7.45 mol H2O

Molar mass of water is the sum of each elements AMU that
2 H + 1 O = H2O

2(1 g/mol) + 1(16/g/mol) = 18 g/mol

use that to convert to grams

7.45 mol H2O * 18 g/mol H2O = 134.18 g H2O

Answer 2

This one actually is shorter. I don't know if they still teach things this way.
1 mole of gas occupies 22.4 liters. at STP. I might have been able to do the question you asked previously that way. It depends. But knowing about PV = nRT doesn't hurt you either.

One
Balance the equation which is done.

Two
Calculate the number of moles of oxygen.
1 mol oxygen needs a volume of 22.4 liters at STP
x mol oxygen occupies a volume of 83.4 Liters (That's a given)

1/x = 22.4/83.4 Cross multiply
83.4 * 1 = 22.4 * x
83.4/22.4 = x
x = 3.72 moles.

Three
Find the mols of water. Here's where you use the balanced equation.
1 mol of oxygen produces 2 mols of water
3.72 mol of oxygen = x mols of water

1/3.72 = 2/x Cross multiply
x = 2 * 3.72
x = 7.44 mols of water.

Four
Find mass of 1 mol of water
2H = 2*1 = 2
O = 1 * 16 = 16
1 mol of water is 18 grams of water.

Five
Find the mass of 7.44 mols

1 mol water = 18 grams
7.44 mol = x Cross multiply
x = 7.33 * 18
x = 134 grams

D <<<< answer