Question

A chemist mixes 24.0 g H2 with 8.0 g N2. Assuming the reaction goes to completion, which reactant should she use to calculate the yield?

3H2 + N2 + 2NH3
A. Either reactant, because each will be completely used up in the reaction
B. Either reactant, because they are in a ratio of 3:1 for H2:N2
C. Hydrogen, because it could produce the greater amount of product
D. Nitrogen, because the ratio of moles is greater than 3:1 for Hz:N2​

Answer 1

D. Nitrogen, because the ratio of moles is greater than 3:1 for H2:N2

Step-by-step explanation:

Hello,

In this case, given the reaction:

`3H_2 + N_2 \rightarrow 2NH_3`

In order to identify the limiting reactant we must compute the moles of ammonia yielded by both reactants at first:

`n_(NH_3)^(by\ H_2)=24.0gH_2*(1molH_2)/(2molH_2)*(2molNH_3)/(3molH_2)=8molNH_3\\ \\n_(NH_3)^(by\ N_2)=8.0gN_2*(1molN_2)/(28molN_2)*(2molNH_3)/(1molN_2)=0.57molNH_3`

Thus, since nitrogen yields the smallest amount of ammonia as it is more heavy than hydrogen and it is in a 3:1 mole ratio for H2:N2, it is the limiting reactant, therefore the answer is D. Nitrogen, because the ratio of moles is greater than 3:1 for H2:N2.

Regards.