Question

2.00g H2 is allowed to react with 9.88g N2, producing 1.57g NH3
What is the theoretical yield in grams for this reaction under the given conditions?
What is the percent yield for this reaction under the given conditions?

Answer 1

Balanced chemical reaction is given by:

$N_2(g)\ \ \ +\ \ 3H_2(g)\ \ \to\ \ \ 2NH_3(g)$

1 mole nitrogen will react completely with 3 mole hydrogen

0.353 mole (or 9.88 g) nitrogen will react with $0.353\times 3=1.059\ mole$ hydrogen = 2.118 g $H_2$

So, here Hydrogen is the limiting reagent

So theretically

3 mole of hydrogen will produce 2 mole ammonia

So, 1 mole(or 2 g) of hydrogen will produce $\frac{2}{3}\ mole$ ammonia

So theoretical yield = $\frac{2}{3}\ mole\ NH_3=\frac{2}{3}\times 17=11.33\ g\ NH_3$

Percentage yield = $\frac{actual \ yield}{theoretical\ yield}\times 100=\frac{1.57}{11.33}\times 17=13.857\ \%$