Question

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

A. What is the theoretical yield in grams for this reaction under the given conditions?
B. What is the percent yield for this reaction under the given conditions?

Answer 1

A.
`m_(NH_3)^(theo) =1.50gNH_3`

B.
`Y=82.2\%`

Step-by-step explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

`N_2+3H_2\rightarrow 2NH_3`

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

`n_(NH_3)^(by \ H_2)=1.12gH_2*(1molH_2)/(2.02gH_2)*(2molNH_3)/(3molH_2)=0.370molNH_3\\\\ n_(NH_3)^(by \ N_2)=1.23gN_2*(1molN_2)/(28.02gN_2)*(2molNH_3)/(1molN_2)=0.0878molNH_3`

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

`m_(NH_3)^(theo)=0.0878mol*(17.04gNH_3)/(1molNH_3) =1.50gNH_3`

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

`Y=(1.23gNH_3)/(1.50gNH_3) *100\%\\\\Y=82.2\%`

Best regards!