Answer to Question #111668 in General Chemistry for sydney

Question

The following reaction is performed: 3H₂ + N₂ → 2NH₃ . You begin this reaction with 10.00 grams of each reactant. What is the excess reactant?

Answer 1

3H₂ + N₂ → 2NH₃

n_H2 = m_H2/Mr_H2 = 10 g / 2 (g/mol) = 5 mol

Per 1 mol: 5/3 = 1,66 mol of H₂

n_N2 = m_N2/Mr_N2 = 10 g / 28 (g/mol) = 0,357 mol. Its per 1 mol.

So, the excess reactant is N_2.