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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
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