Question

Nitrogen gas (N2) reacts with excess hydrogen gas (H2) to produce ammonia (NH3). What is the percent yield of ammonia if the actual yield is 1.03 moles and the theoretical yield is 19.7 grams?
a. 17.5%
b. 35.0%
c. 52.3%
d. 89.0%

Answer 1

a. 17.5%$\frOKUU+ac{x}{y}$≡⇒∛Ф⇔
HHUGT,NHY

Answer 2

Answer : The percent yield of ammonia is, 89.0 %

Explanation :  Given,

Actual yield of $NH_3$ = 1.03 mole

Theoretical yield of $NH_3$ = 19.7 g

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the actual yield of $NH_3$ in terms of mass.

$\text{Mass of }NH_3=\text{Moles of }NH_3\times \text{Molar mass of }NH_3$

$\text{Mass of }NH_3=(1.03mole)\times (17g/mole)=17.51g$

Now we have to calculate the percent yield of $NH_3$

$\%\text{ yield of }NH_3=\frac{\text{Actual yield of }NH_3}{\text{Theoretical yield of }NH_3}\times 100=\frac{17.51g}{19.7g}\times 100=88.88\%\approx 89.0\%$

Therefore, the percent yield of ammonia is, 89.0 %