Asked by David
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 :)
Completly confused, thank you very much for your help :)
Answers
Answered by
DrBob222
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).
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).
Answered by
Trey
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?
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?
Answered by
hello internet stranger
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.
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.
Answered by
hello internet stranger
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.
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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