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

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

The answer to the question is d

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