In the reaction of N2 and H2 to produce NH3, how many moles of H2 will produce 37.9 grams NH3 if sufficient N2 is present? Do not enter units with your answer.
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To solve this problem, we need to use the balanced chemical equation for the reaction of N2 and H2 to produce NH3, which is:
N2 + 3H2 → 2NH3
From the equation, we can see that the stoichiometric ratio between N2 and H2 is 1:3, which means that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Now, we need to find how many moles of H2 are required to produce 37.9 grams of NH3.
First, we need to convert grams of NH3 to moles using the molar mass of NH3:
Molar mass of NH3 = 14.01 g/mol (N) + 1.01 g/mol (H) x 3 = 17.03 g/mol
37.9 g NH3 x (1 mol NH3 / 17.03 g NH3) ≈ 2.23 mol NH3
According to the stoichiometric ratio, 3 moles of H2 are required to produce 2 moles of NH3.
So, using a proportion, we can find the number of moles of H2:
(2.23 mol NH3) x (3 mol H2 / 2 mol NH3) ≈ 3.35 mol H2
Therefore, approximately 3.35 moles of H2 are needed to produce 37.9 grams of NH3, assuming sufficient N2 is present.
To determine the number of moles of H2 required to produce 37.9 grams of NH3, we need to use the balanced chemical equation for the reaction:
N2 + 3H2 -> 2NH3
From the balanced equation, we can see that 3 moles of H2 are required to produce 2 moles of NH3.
First, we need to find the molar mass of NH3:
N + 3H = 14.01 g/mol + (3 × 1.01 g/mol) = 17.03 g/mol
Next, we can use the molar mass of NH3 to calculate the number of moles of NH3:
37.9 g NH3 × (1 mol NH3 / 17.03 g NH3) = 2.23 mol NH3
Since the stoichiometric ratio between H2 and NH3 is 3:2, we can now calculate the number of moles of H2:
2.23 mol NH3 × (3 mol H2 / 2 mol NH3) = 3.35 mol H2
Therefore, 3.35 moles of H2 will produce 37.9 grams of NH3 when sufficient N2 is present.