# In the reaction N2 + 3H2 -> 2NH3, how many grams of NH3 gas would be formed when H2 gas, having a volume of 11.2 liters at STP, reacts with excess N2 gas?

Completly confused, thank you very much for your help :)

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1. Given the following chemical reaction:

N2 + 3H2 → 2NH3

How many grams of NH3 would be produced if you started out with 56 grams of nitrogen gas and 10 grams of hydrogen gas?

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2. 2NH3 mass = 17*2 = 34 g
3H2 mass = 6 g
11.2 liters H2 = 11.2/22.4 = 0.5 mole = 0.5 * 2 = 1 g
34 x = 6 1
x = 5.667

5.7 grams NH3 is required for 11.2 liters H2.

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3. edited: last one didn't load divison signs

2NH3 mass = 17*2 = 34 g
3H2 mass = 6 g
11.2 liters H2 = 11.2/22.4 = 0.5 mole = 0.5 * 2 = 1 g
34/x = 6/1
X = 5.667

5.7 grams NH3 is required for 11.2 liters H2.

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4. Here is a link that will show you how to work stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html

There is a shortcut that can be used when all of the reactants and products are gases. You may consider volumes as moles. That way it looks like this.
11.2 L H2 x (2 moles NH3/3 mole H2) = 11.2 x (2/3) = xx L NH3.
Then convert xx L NH3 to grams remembering that 22.4 L NH3 has a mass of 17 grams (1 mole).

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