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ammomia is produced from the reaction of nitrogen according to the following balance N2(g)+ 3H2(g)---- 2NH3(g)?
what is the maximum mass of ammonia that can produced from a mixture of 1.00 *10^3 g N2 and 5.00 *10^2 g H2?
1 Answer
- Roger the MoleLv 72 years ago
(1000 g N2) / (28.01344 g N2/mol) = 35.697 mol N2
(500 g H2) / (2.01588 g H2/mol) = 248.031 mo H2
35.697 moles of N2 would react completely with 35.697 x (3/1) = 107.091 moles of H2, but there is more H2 present than that, so H2 is in excess and N2 is the limiting reactant.
(35.697 mol N2) x (2 mol HN3 / 1 mol N2) x (17.03056 g NH3/mol) = 1.22 x 10^3 g NH3
