Q4.109P - Chemistry: Molecular Nature Of Matter And Change

Question

Ammonia is produced by the millions of tons annually for use as a fertilizer. It is commonly made from N₂ and H₂ by the Haber process. Because the reaction reaches equilibrium before going completely to product, the stoichiometric amount of ammonia is not obtained. At a particular temperature and pressure, 10.0 g of H₂ reacts with 20.0 g of N₂ to form ammonia. When equilibrium is reached, 15.0 g of NH₃ has formed. 

 

(a) Calculate the percent yield. 

 

(b) How many moles of N₂ and H₂ are present at equilibrium?

Step-by-Step Solution

Answer

Answer of subpart (a):

 

Answer: You need to calculate the percent yield.

Answer of subpart (b):

 

Answer: You need to calculate how many moles of N₂ and H₂ are present at equilibrium.

Step 1: Balanced equation

Nitrogen reacts with hydrogen and produces ammonia. The balanced equation is like

 

 ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Step 2: Calculation of moles of N 2

According to the question,

 

Mass of N₂ = 20g

 

Molecular mass of N₂ = 28g/mol

 

Again you know,

 $\text{moles = }\frac{\text{mass }\left(\text{given}\right)}{\text{mass }\left(\text{molar}\right)}$

 

Moles of N₂

 $\begin{array}{c}=\frac{20}{28}\left(\mathrm{mol}\right)\\ =0.714\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of N₂ are 0.714mol.

Step 3: Calculation of moles of H 2

According to the question,

 

Mass of H₂ = 10g

 

Molecular mass of H₂ = 2.016g/mol

 

Again you know,

$\mathrm{moles}=\frac{\mathrm{mass}\left(\mathrm{given}\right)}{\mathrm{mass}\left(\mathrm{molar}\right)}$

Moles of H₂

 $\begin{array}{l}=\frac{10}{2.016}\left(\mathrm{mol}\right)\\ =4.96\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of H₂ are 4.96mol.

Step 4: Calculation of moles of NH 3 when N 2 is limiting reagent

According to the balanced equation

 

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

 

1 mole of N₂ produces 2 moles NH₃

 

Now, moles of NH₃

 $\begin{array}{l}=0.714\times \frac{2}{1}\left(\mathrm{mol}\right)\\ =1.428\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of NH₃ when N₂ is limiting reagent are 1.428mol.

Step 5: Calculation of moles of NH 3 when H 2 is limiting reagent

According to the balanced equation

 

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

 

3 moles of H₂ produces 2 moles NH₃

 

Now, moles of NH₃

 $\begin{array}{l}\text{=4.96×}\frac{\text{2}}{\text{3}}\left(\text{mol}\right)\\ \text{=3.307}\left(\text{mol}\right)\end{array}$

 

Hence, moles of NH₃ when H₂ is limiting reagent are 3.307mol.

Step 6: Conclusion

N₂ is the limiting reagent in this reaction as moles of NH₃ is less in terms of N₂.

Step 7: Calculation of mass of NH 3

Moles of NH₃ = 1.428moles

 

Molecular mass of NH₃ = 17.024g/mol

 

Again you know,

 $\mathrm{mass}=\mathrm{moles}\times \mathrm{mass}\left(\mathrm{molar}\right)$

 

Mass of NH₃ 

 $\begin{array}{l}=1.428\times 17.024\left(\mathrm{g}\right)\\ =24.31\left(\mathrm{g}\right)\end{array}$

 

Hence, mass of NH₃ is 24.31g.

Step 8: Percent yield

As you know, percent yield

 

 $\begin{array}{l}=\frac{\mathrm{yield}\left(\mathrm{actual}\right)}{\mathrm{yield}\left(\mathrm{theoritical}\right)}\times 100\mathrm{\%}\\ =\frac{15}{24.31}\times 100\mathrm{\%}\\ =61.7\mathrm{\%}\end{array}$

Step 9: Conclusion

Hence, the percent yield is 61.7%.

Answer of subpart (b):

 

Answer: You need to calculate how many moles of N₂ and H₂ are present at equilibrium.

Step 10: Calculation of moles of NH 3

According to the question,

 

Mass of NH₃ = 15g

 

Molecular mass of NH₃ = 17.024g/mol

 

Again you know,

 $\text{moles=}\frac{\text{mass}\left(\text{given}\right)}{\text{mass}\left(\text{molar}\right)}$

 

Moles of NH₃

 

 $\begin{array}{l}=\frac{15}{17.024}\left(\mathrm{mol}\right)\\ =0.88\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of NH₃ are 0.88mol.

Step 11: Calculation of moles of N 2 produce 15g of NH 3

According to the balanced equation

 

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

 

1 mole of N₂ produces 2 moles NH₃

 

Now, moles of N₂

 $\begin{array}{l}=0.88\times \frac{1}{2}\left(\mathrm{mol}\right)\\ =0.44\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of N₂ produce 15g NH₃ are 0.44mol.

Step 12: Calculation of moles of H 2 produce 15g NH 3

According to the balanced equation

 

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

 

3 moles of H₂ produces 2 moles NH₃

 

Now, moles of H₂

 $\begin{array}{l}=0.88\times \frac{3}{2}\left(\mathrm{mol}\right)\\ =1.32\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of H₂ produce 15g NH₃ are 1.32mol.

Step 13: Calculation of moles of N 2 at equilibrium

As you know,

 

Moles at equilibrium = initial moles – reacted moles

 

Now, initial moles of N₂ = 0.714mol

 

Now, moles of N₂ at equilibrium

 

 $\begin{array}{l}=(0.714-0.44)\left(\mathrm{mol}\right)\\ =0.274\left(\mathrm{mol}\right)\end{array}$

Step 14: Calculation of moles of H 2 at equilibrium

As you know,

 

Moles at equilibrium = initial moles – reacted moles

 

Now, initial moles of H₂ = 4.96mol

 

Now, moles of H₂ at equilibrium

 

 $\begin{array}{l}=(4.96-1.32)\left(\mathrm{mol}\right)\\ =3.64\left(\mathrm{mol}\right)\end{array}$

Step 15: Conclusion

Moles of N₂ and H₂ at equilibrium are 0.274mol and 3.64mol respectively.