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N2 + 3H2 -> 2NH3

The stoichiometric equation (or balanced equation) for the formation of ammonia

from this we can read off the mole ratio between hydrogen and ammonia;

3M H2 needed to produce 2M NH3

times each by 9 (so the ratio remains the same and 18M NH3 is formed)

27M H2 needed to produce 18M NH3

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Q: How many moles of hydrogen are required to produce 18.00 moles of ammonia?
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How many moles of ammonia are produced from 12.0 moles of hydrogen?

N2 + 3H2 -----> 2NH3 so 3 moles of hydrogen produce 2 moles of ammonia. Therefore 12.0 moles of hydrogen will produce 8 moles of ammonia.


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That amount of ammonia contains two moles of hydrogen gas. One mole of hydrogen gas weighs 2.016 grams. Therfore 3.75 grams of ammonia contains two moles of hydrogen.


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To form ammonia, reaction is N(2) + 3H(2) ---> 2NH(3) + H(2)O. As you can see for 1 mole of nitrogen three moles of hydrogen is required. Hence for your question, 1.13 moles nitrogen is required.


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How much hydrogen would be required to produce 7.5 mol of water?

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N2 + 3H2 ---> 2NH3 so 3 moles of Hydrogen produces 2 moles of ammonia. Thus 1.8 moles will produce 1.8/3 x 2 = 1.2 moles of ammonia.


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The reaction of nitrogen with hydrogen to form ammonia is: N2 +3H2 = 2NH3 Therefore to make 10 moles of ammonia you need 5 moles N2 and 15 moles H2


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To form ammonia, balanced reaction is N(2) + 3H(2) ---> 2NH(3) + H(2)O. As you can see for 1 mole of nitrogen three moles of hydrogen is required. Hence for your question, 3 moles nitrogen is required to satisfy the ratio.


How many molecules of ammonia can be formed from the reaction of 2 molecules of Nitrogen gas and 6 molecules of hydrogen gas?

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