This is based on calculations too. It contains 18 hydrogen moles.
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 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.
this is a easy one. There are only 0.04166 moles.
The balanced chemical equation for the reaction between H2 and NH3 is: 3H2 + N2 -> 2NH3 From the balanced equation, we can see that 3 moles of H2 produce 2 moles of NH3. Therefore, 1.8 moles of H2 will produce 1.2 moles of NH3.
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 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
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.
To calculate the moles of hydrogen needed to produce 68 grams of ammonia (NH₃), we start with the balanced chemical equation for the synthesis of ammonia: N₂ + 3H₂ → 2NH₃. The molar mass of ammonia is approximately 17 g/mol, so 68 grams of NH₃ corresponds to 68 g / 17 g/mol = 4 moles of NH₃. Since 3 moles of hydrogen are required for every 2 moles of ammonia, the moles of hydrogen needed is (4 moles NH₃) × (3 moles H₂ / 2 moles NH₃) = 6 moles of H₂. Therefore, 6 moles of hydrogen must react to produce 68 grams of ammonia.
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
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 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.
The balanced chemical equation for the formation of water from ammonia is: 2 NH₃ + 3 O₂ → 2 H₂O + N₂. According to the equation, 2 moles of ammonia produce 2 moles of water. Therefore, to produce 13.8 moles of water, you would require 13.8 moles of ammonia, since the ratio is 1:1.
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.
Three moles of nitrogen are required to produce 2 moles of ammonia according to the balanced chemical reaction for ammonia synthesis. Therefore, 27 moles of nitrogen are required to produce 18 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.
this is a easy one. There are only 0.04166 moles.
The balanced chemical equation for the reaction between ammonia (NH3) and oxygen gas (O2) is 4 NH3 + 5 O2 → 4 NO + 6 H2O. This means that 5 moles of O2 are needed to react with 4 moles of NH3. With 10.0 moles of NH3, you would need 12.5 moles of O2 (10.0 moles NH3 x 5 moles O2 / 4 moles NH3).