Question

PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will produce ammonia (NH3). Hydrogen and nitrogen gases are reacted to produce the ammonia. For the first batch of ammonia production, 475 g of nitrogen is reacted with excess hydrogen, and 397 g of ammonia are produced.
• Write the balanced equation for the formation of ammonia from hydrogen and nitrogen.
3H2+N2-->2NH3

• Calculate the theoretical yield of ammonia. Work must be shown to earn credit.



• Calculate the percent yield for the ammonia production. Work must be shown to earn credit.

Answer 1

Answer :

Part (a) :  $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part (b) :  Theoretical yield of $NH_3$ = 440.96 g

Part (c) : Percentage yield of ammonia is 90.03 %

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

Part (a) :The balanced chemical reaction will be:

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

Part (b) :To calculate the moles of $N_2$:

$\text{Moles of }N_2=\frac{\text {Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given balanced equation:

1 mole of $N_2$ gas produces 2 moles of $NH_3$

16.96 moles of $N_2$ gas produces $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

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

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

Part (c) : Percentage yield :

$\% \text{ yield of }NH_3=\frac{\text {Experimental yield of NH_3}}\text {Theoretical yield of }NH_3\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %