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.
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
The balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia is: 3H2 + N2 -> 2NH3 From the equation, it can be seen that 3 moles of hydrogen react to produce 2 moles of ammonia. Therefore, 18 moles of hydrogen can produce (2/3) x 18 = 12 moles of ammonia.
3H2 + N2 <------> 2NH3 is the balanced equation for Hydrogen and Nitrogen making ammonia. 3 moles of H2 produces two moles of ammonia and thus to make 6 moles requires 9 moles of Hydrogen.
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 balanced equation for the reaction is: 3H2 + N2 -> 2NH3 From the balanced equation, we can see that 3 moles of hydrogen are needed to react completely with 1 mole of nitrogen. So if there are 3 moles of nitrogen, you would need 9 moles of hydrogen to react completely.
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.
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.
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 balanced chemical equation for the formation of ammonia from nitrogen and hydrogen is N2 + 3H2 → 2NH3. From the equation, it can be seen that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Calculate the moles of nitrogen and hydrogen provided, determine the limiting reactant, and then use stoichiometry to find the grams of ammonia that can be produced.
Since ammonia has a chemical formula of NH3, it contains one mole of nitrogen and three moles of hydrogen per mole of ammonia. Therefore, 3 moles of ammonia contain 3 moles of nitrogen and 9 moles of hydrogen atoms.
N2 + 3H2 -> 2NH3 3 moles hydrogen gas. You should know that because of the formula of 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
The balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia is: 3H2 + N2 -> 2NH3 From the equation, it can be seen that 3 moles of hydrogen react to produce 2 moles of ammonia. Therefore, 18 moles of hydrogen can produce (2/3) x 18 = 12 moles of ammonia.
To produce 525 grams of ammonia (NH3), you would need 25 moles of ammonia. Since the balanced chemical equation for the reaction between hydrogen and nitrogen to form ammonia is 3H2 + N2 -> 2NH3, you would need 75 moles of hydrogen molecules (H2) to produce 525 grams of ammonia. This is equivalent to 4,500 molecules of hydrogen.
3H2 + N2 <------> 2NH3 is the balanced equation for Hydrogen and Nitrogen making ammonia. 3 moles of H2 produces two moles of ammonia and thus to make 6 moles requires 9 moles of Hydrogen.
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.