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What mass of NH3 you prepared from the reaction of 10g of N2 with excess H2?
Wiki User
∙ 12y ago
I had to do this problem on a sample exam I got from the University of South Carolina. It has to do with a stoichiometry reaction. If I got the answer wrong, I apologize but this is how I did it. I'll use 2 approaches so you can understand all the concepts going into it. And forget about the "excess H_2" so it doesn't throw you off.
First write a balanced equation of the reactants and product:
N_2 + 3H_2 -----> 2NH_3
N_2 = 10g
10g of N_2 x [1 mol N_2 / (14.0067x2)] x (3 mol H_2 / 1 mol N_2) x [(1.0079x2) / 1 mol H_2]
so,
10 x (1 / 28.0134) x (3) x (2.0158), is
10 x (.03569) x (3) x (2.0158) = 2.158 grams of H_2
Since we have 10g of N_2, we can add the two numbers and find out the mass of ammonia. 10g + 2.158 = 12.158g, or 12 g NH_3.
Similarly we can substitute the ammonia (NH_3) for the H_2 in the equation above and get the same answer.
10g of N_2 x [1 mol N_2 / (14.0067x2)] x (2 mol NH_3 / 1 mol N_2) x [(14.0067 + (1.0079 x 3)) / 1 mol NH_3]
so,
10 x (1 / 28.0134) x (2) x (17.0304), is
10 x (.03569) x (2) x (17.0304) = 12.158 g, or 12 g NH_3
Wiki User
∙ 12y ago
By stoichiometry, every 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, 10g of N2 corresponds to 0.357 moles. Since there is excess H2, all 10g of N2 would react to produce NH3, which would be 0.476 moles (0.357 moles of N2 * 2 moles of NH3 / 1 mole of N2). The mass of NH3 produced would be 9.07g (0.476 moles * 17.03g/mol).
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