Question

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?

Answer 1

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 = ?