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.
The answer is 97,66 moles.
CO2 + H2 -> CO + H2O one to one here 30.6 moles H2O (1 mole H2/1 mole H2O) = 30.6 moles Hydrogen gas needed
The combustion reaction of ethane (C2H6) with oxygen produces carbon dioxide and water. The balanced equation for the combustion of ethane is: [ 2 \text{C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} ] From the equation, 2 moles of ethane produce 6 moles of water. Therefore, 4 moles of ethane will produce ( \frac{6}{2} \times 4 = 12 ) moles of water.
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.
1 mole
methane has one carbon atom and 4 hydrogen atom.....so one mole of methane contains one mole of carbon and 4 moles of hydroge......for simplicity...one mole of water has two moles of hydrogen and one mole of oxygen....
The answer is 97,66 moles.
This is based on calculations too. It contains 18 hydrogen moles.
To find the number of hydrogen atoms in 90 amu of ethane (C2H6), use the molar mass of ethane to determine the number of moles present. Next, use the molecular formula of ethane to calculate the number of hydrogen atoms in one mole, and then multiply by the number of moles present to find the total number of hydrogen atoms.
CO2 + H2 -> CO + H2O one to one here 30.6 moles H2O (1 mole H2/1 mole H2O) = 30.6 moles Hydrogen gas needed
Approx. 4 moles.
2 moles of sodium will produce 1 mole of hydrogen gas according to the chemical equation 2Na + 2H2O → 2NaOH + H2. The molar mass of sodium is 23 g/mol and of hydrogen gas is 2 g/mol. Thus, 2 moles of sodium is 46 grams (2 moles * 23 g/mol), which will produce 2 moles of hydrogen gas.
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 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.
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.
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