91Ó°ÊÓ

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

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

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

Hence, moles of N₂ are 0.714mol.

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.

According to the balanced equation

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

1 mole of N₂ produces 2 moles NH₃

Now, moles of NH₃

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

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

According to the balanced equation

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

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.

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

Moles of NH₃ = 1.428moles

Molecular mass of NH₃ = 17.024g/mol

Again you know,

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

Mass of NH₃

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

Hence, mass of NH₃ is 24.31g.

As you know, percent yield

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

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.

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.

According to the balanced equation

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

1 mole of N₂ produces 2 moles NH₃

Now, moles of N₂

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

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

According to the balanced equation

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

3 moles of H₂ produces 2 moles NH₃

Now, moles of H₂

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

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

As you know,

Moles at equilibrium = initial moles – reacted moles

Now, initial moles of N₂ = 0.714mol

Now, moles of N₂ at equilibrium

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

As you know,

Moles at equilibrium = initial moles – reacted moles

Now, initial moles of H₂ = 4.96mol

Now, moles of H₂ at equilibrium

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

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