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.
Pure methane cannot contain ammonia molecules. These are two different compounds
well if thats 75.0g there would be 37.13 moles
2NH3 ----> 3H2 + N2 so 2 moles of ammonia produce 1 of Nitrogen. So to produce 3.5 moles of nitrogen would require 2 x3.5 = 7 moles.
8,038 moles of ammonia were produced.
This is based on calculations too. It contains 18 hydrogen moles.
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.
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)
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.
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.
This is based on calculations too. It contains 18 hydrogen moles.
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.
1 mole of nitrogen makes 3 of ammania so.. moles of ammonia=34/17=2 moles as 1 mole of nitrogen makes 3 of ammania so.. 2/3 x 2 x14 = 18.67 g
The chemical equation is N2 + 3H2 -> 2NH3 So reacting 2 moles of N2 will produce 4 moles ammonia.
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.
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)
10 moles of nitrogen dioxide are needed to react with 5,0 moles of water.
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