18 mol of NH3 (ammonia) contains 18 mol of N atoms.
1 mole of N2 (nitrogen gas) contains 2 mol N atoms.
so 9 mol N2 is used to produce 18 mol NH3.
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
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.
Each mole of ammonia requires one mole of nitrogen atoms. However, the nitrogen in the air occurs as diatomic molecules; therefore, only one-half mole of molecular nitrogen is required for each mole of ammonia.
The chemical equation is N2 + 3H2 -> 2NH3 So reacting 2 moles of N2 will produce 4 moles ammonia.
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
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.
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.
9H2 + 3N2 ------> 6NH3 3 moles of N2 would be required.
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.
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
8,038 moles of ammonia were produced.
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.
Each mole of ammonia requires one mole of nitrogen atoms. However, the nitrogen in the air occurs as diatomic molecules; therefore, only one-half mole of molecular nitrogen is required for each mole of ammonia.
The chemical equation is N2 + 3H2 -> 2NH3 So reacting 2 moles of N2 will produce 4 moles ammonia.
There are two elements that make up Ammonia (NH3) … One mole of Nitrogen (N) plus three moles of Hydrogen (H) react to produce one mole of Ammonia (NH3)
Assuming that you mean the reaction of nitrogen and hydrogen to form ammonia N2 + 3H2 -> 2NH3 1 mole of nitrogen forms 2 moles of ammonia- so 4.08 l of nitrogen will be consumed to form 8.16 moles of ammonia. This assumes both are pretty ideal gases which is a reasonabale approximation.
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