Question

Consider the reaction:

N2(g) + 3H2(g) → 2NH3(g)

Given an initial mass of 14.37 g N2, an excess of H2, and assuming that all of the reactant is converted to product(s), and none is lost, calculate the mass (g) of NH3 produced by the reaction.

Answer 1

N2(g) + 3H2(g) → 2NH3(g)

m(NH3)=14.37(g)*2*17(g/mol)/14(g/mol)*2=17.45(g)