This is not a common reaction at standard temperature and pressure.
The molar mass of aluminum is 27 g/mol. From the balanced equation, 2 moles of aluminum produce 3 moles of hydrogen. Therefore, 1.5 g of hydrogen is produced by 1 g of aluminum. Hence, the mass of aluminum required to produce 1.5 g of hydrogen gas is 1.5 g.
To produce 1 mole of water, you need 2 moles of hydrogen. Therefore, to produce 7.4 moles of water, you would need 2 * 7.4 = 14.8 moles of hydrogen.
This is a chemical equation describing the reaction between hydrochloric acid and aluminum to form aluminum chloride and hydrogen gas. Written with formatting, the chemical equation looks like: 6 HCl + 2 Al --> 2 AlCl3 + 3 H2
To produce 1 mol of water, 2 mol of hydrogen is needed. Therefore, to produce 7.5 mol of water, you would need 15 mol 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 molar mass of aluminum is 27 g/mol. From the balanced equation, 2 moles of aluminum produce 3 moles of hydrogen. Therefore, 1.5 g of hydrogen is produced by 1 g of aluminum. Hence, the mass of aluminum required to produce 1.5 g of hydrogen gas is 1.5 g.
To produce 1 mole of water, you need 2 moles of hydrogen. Therefore, to produce 7.4 moles of water, you would need 2 * 7.4 = 14.8 moles of hydrogen.
This is a chemical equation describing the reaction between hydrochloric acid and aluminum to form aluminum chloride and hydrogen gas. Written with formatting, the chemical equation looks like: 6 HCl + 2 Al --> 2 AlCl3 + 3 H2
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.
When 4 moles of aluminum react with an excess of chlorine gas, 4 moles of aluminum chloride are produced. This is because the balanced chemical equation for the reaction is: 2Al + 3Cl2 -> 2AlCl3 This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride, so 4 moles of aluminum will produce 4 moles of aluminum chloride.
To produce 1 mol of water, 2 mol of hydrogen is needed. Therefore, to produce 7.5 mol of water, you would need 15 mol 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.
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.
3.2 moles of water (H2O)
20 moles of Al will react with water to form 60 moles of H2 and 30 moles of O2, because of the stoichiometry 2:1 in getting H2 en O2 from water, and Al needing an oxidationstate of 3+, so 1 mole is oxidized by 3 moles of H2O.
To produce ethane (C₂H₆), the balanced chemical equation is: ( 2 \text{C} + 3 \text{H}_2 \rightarrow \text{C}_2\text{H}_6 ). This indicates that 3 moles of hydrogen (H₂) are required for every 1 mole of ethane produced. Therefore, to produce 13.78 moles of ethane, you would need ( 3 \times 13.78 = 41.34 ) moles of hydrogen.