Ammonia is produced when Nitrogen gas reacts with Hydrogen gas, as shown by the equation below:

N2 + 3H2 --> 2NH3

If the percentage yield is 90%, what would be your experimental yield of NH3 if 145.8g of N2 react?

2) 2Mg + O2 --> 2MgO

A) If 50g of Mg reacts with 50g of O2, what is the limiting reactants?

B) How much excess reactant is left over

C) How much product is produced?

1 answer

mols N2 = 145.8g/molar mass N2 = ?
Convert mols N2 to mols NH3 using the coefficients in the balanced equation.
Now convert mols NH3 to g. g = mols x molar mass. This is the theoretical yield. If the actual yield is only 90%, then g NH3 x 0.90 = ?
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