Question

3H2(g)+N2(g)→2NH3(g) 1.36 g H2 is allowed to react with 9.51 g N2, producing 1.06 g NH3 1.) What is the theoretical yield in grams for this reaction under the given conditions? 2.)What is the percent yield for this reaction under the given conditions?

Answer 1

`m_(NH_3)=7.71gNH_3`

`Y=13.8\%`

Step-by-step explanation:

Hello,

In this case, given the chemical reaction and the initial amounts, we first identify the limiting reactant by verifying the reactant yielding the smallest amount of product:

`n_(NH_3)^(By\ H_2)=1.36gH_2*(1molH_2)/(2gH_2)*(2molNH_3)/(3molH_2)=0.453molNH_3\\ \\n_(NH_3)^(By\ N_2)=9.51gN_2*(1molNH_2)/(28gN_2)*(2molNH_3)/(1molN_2)=0.679molNH_3`

Thus, since the hydrogen yields the least amount of ammonia, it is the limiting reactant, which means the theoretical yield of ammonia is:

`m_(NH_3)=0.453molNH_3*(17gNH_3)/(1molNH_3)\\ \\m_(NH_3)=7.71gNH_3`

Finally, the percent yield for a 1.06-g actually yielded amount turns out:

`Y=(1.06gNH_3)/(7.71gNH_3)*100\%\\ \\Y=13.8\%`

Best regards.