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0. Hydrogen doesn't "reackt" to form Nitrogen Monoxide.
The reaction to form nitrogen dioxide using nitric oxide is; 2NO(g) + O2(g) -> 2NO2(g) As the stoichiometry between the substances are 1:1, 1.35 moles of nitrogen monoxide is needed.
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.
Mole Ratio
Nitrogen and hydrogen don't form ionic compounds. they form only covalent compounds as in ammonia (NH3) or hydrazine (H2N-NH2) etc
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
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.
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.
0. Hydrogen doesn't "reackt" to form Nitrogen Monoxide.
The reaction to form nitrogen dioxide using nitric oxide is; 2NO(g) + O2(g) -> 2NO2(g) As the stoichiometry between the substances are 1:1, 1.35 moles of nitrogen monoxide is needed.
0.1 mols of N2 * 3 mols of H3/1 mol of N2In other words, for 0.1 mols of N2 times 3 mols of H3 for ever mol of N2.I order to find the other numbers you are going to need a balanced equation:N2+3H2→ 2NH3So you would have 0.3 mols.
First you have to find the limiting reactant. You have .3 moles of nitrogen and .6 moles of hydrogen, but you don't know which one is going to run out first.In any of these stoichiometry problems, you need to write down the formula:N2 + 3H2 → 2NH3Take both nitrogen and hydrogen and figure out how much ammonia is made alone..6 moles Hydrogen ÷ 3 moles hydrogen × 2 moles ammonia = .4 moles ammonia made.3 moles Nitrogen ÷ 1 mole nitrogen × 2 mole ammonia = .6 moles ammonia madeNow you figured out that hydrogen is the limiting reactant and the nitrogen is the excess because less ammonia is made using hydrogen. This measurement is what you will be using for the rest of the problem.Take the limiting reactant and use stoichiometry to find how much ammonia can be made.You could start with .6 moles of hydrogen and do the same conversion as above, but add the step of converting to grams. Or, since you already found out that .4 moles ammonia is made, just convert it to grams. The molecular mass of ammonia is 17.0 grams..4 moles ammonia × 17.0 grams = 6.8 grams ammonia
Balanced equation always and always first! N2 + 3H2 >> 2NH3 As you see the dimensional analysis shows three mols H2 to one mol N2. ( mols are always coefficients ) So, three times two is 6.
Nitrogen Fixation
The synthesis reaction is 2 H2 + O2 = 2 H2O. Every two moles of hydrogen reacts with one mole of oxygen to make two moles of water. Then 30.0 grams of water is 1.67 moles, and 1.67 moles of H2 has a mass of 3.37 grams. 25.0 grams of O2 is .781 moles, so 1.562 moles of H2 are needed, or 3.15 grams.
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.
nitrogen gas and hydrogen gas