Given 40 g of N2 and 80 g H2 to produce NH3.

Question

Provide a balance equation. CalcuIate for the theoretical and percentage yield. Determine the limiting reactant, excess reactant, amount in excess for your reaction.

Answer 1

The balanced equation is N2 + 3 H2 --> 2 NH3

First you need to start by finding the molar mass of H2 and N2

H2= (1.01*2)= 2.02g

N2= (14*2)= 28g

NH3= 14 + (1.01*3)= 17.03g

Give the 40g of N2 and 80g of H2 we can calculate the limiting reactant as follows:

80g H2 x (1 mol H2/ 2.02g) x (1 mol N2/ 3 mol H2) x (28g N2/1 mol N2)= 370g N2

40g N2 x (1 mol N2/ 28g N2) x (3 mol H2/ 1 mol N2) x (2g H2/ 1 mol H2)= 8.57g H2

Here we see that the N2 is the limiting reactant because we are only given 40g vs the 370g we need. H2 is the excess reactant because we are given 80g and we only needed 8.57g

Excess amount= 80-8.57= 71.43g H2

To determine the theoretical yield, we use the mass of the limiting reactant:

40g N2 x (1 mol N2/ 28g N2) x (2 mol NH3/ 1 mol N2) x (17.03g NH3/ 1 mol NH3)= 48.66g NH3

The actual yield should have been provided to you and you could use that to find the percent yield using this equation:

percent yield= (actual yield/ theoretical yield) x 100%