Answer:
[tex]m_{NH_3}=7.71gNH_3[/tex]
[tex]Y=13.8\%[/tex]
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:
[tex]n_{NH_3}^{By\ H_2}=1.36gH_2*\frac{1molH_2}{2gH_2}*\frac{2molNH_3}{3molH_2}=0.453molNH_3\\ \\n_{NH_3}^{By\ N_2}=9.51gN_2*\frac{1molNH_2}{28gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3[/tex]
Thus, since the hydrogen yields the least amount of ammonia, it is the limiting reactant, which means the theoretical yield of ammonia is:
[tex]m_{NH_3}=0.453molNH_3*\frac{17gNH_3}{1molNH_3}\\ \\m_{NH_3}=7.71gNH_3[/tex]
Finally, the percent yield for a 1.06-g actually yielded amount turns out:
[tex]Y=\frac{1.06gNH_3}{7.71gNH_3}*100\%\\ \\Y=13.8\%[/tex]
Best regards.
