Chapter 13, Problem 42A

Interpretation Introduction

Interpretation: -Volume of ammonia gas is calculated using given variable values given

Concept Introduction: - For calculating volume of ammonia, first of all we have to calculate the number of moles of magnesium nitride is used.

There is a relation between number of moles and volume which is expressed by using Avogadro’s law.

Volume of ammonia gas $5.034L$

There is a relation between volume of a gas and number of molecules of gas present in it, this relation is expressed by Avogadro’s law.

For calculating volume, we use formula of ideal gas

$\mathrm{PV}=nRT$

According to Avogadro’s law, volume occupied by 1 mole every substance is 22.4L at NTP weigh’s equal to the molecular mass and contains Avogadro’s number

  $6.023\times {10}^{23}$ of particles

  $\begin{array}{l}no.ofmolesof=\frac{givenmass}{molarmass}\\ =\frac{5.53g}{44.1g/mol}\\ =0.125moles\end{array}$

  ${\mathrm{C}}_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

According to stoichiometry I mole of propane requires 5 moles of Oxygen. Requires

= 0.125 $\times$ 5 = 0.625 moles of oxygen.

According to the Ideal gas equation

  $\mathrm{PV}=nRT$

  $\mathrm{V}=\frac{nRT}{P}$

From this volume of a gas is calculated.

The equation is given by:

  ${\mathrm{Mg}}_3{N}_2(s)+3H_{2}O(l)\to 3MgO(s)+2N{H}_3(g)$

Molar mass of ${\mathrm{Mg}}_3{N}_2$ is calculated as:

  $(24.305\times 3)+(14.0067\times 2)=100.9284g/mol$

So. The number of moles of ${\mathrm{Mg}}_3{N}_2$ present in

no. of moles $\text{=}\frac{\text{mass}}{\text{molar mass}}$

  $\begin{array}{l}=\frac{10.3g}{100.9284g/mol}\\ =0.1021mol\end{array}$

From the above equation one mole of ${\mathrm{Mg}}_3{N}_2$ will produce 2 moles of ammonia on reaction with water

Hence, the number of moles of ammonia $(N{H}_3)$ evolved.

  $\begin{array}{l}=0.1021\times 2mol\\ =0.2042mol\end{array}$

Volume of ammonia =?

  $\mathrm{PV}=nRT$

  $\mathrm{V}=\frac{nRT}{P}$

  $\begin{array}{l}P=\frac{752}{760}=0.989\\ T=24+273=299\\ V=\frac{0.2042\times 0.0821\times 297}{0.989}\\ V=\frac{4.9791}{0.989}=5.034L\end{array}$

Volume of ammonia is $5.034L$

Volume of ammonia gas $5.034L$

Answer to Problem 42A

Volume of ammonia gas $5.034L$

Explanation of Solution

There is a relation between volume of a gas and number of molecules of gas present in it, this relation is expressed by Avogadro’s law.

For calculating volume, we use formula of ideal gas

$\mathrm{PV}=nRT$

According to Avogadro’s law, volume occupied by 1 mole every substance is 22.4L at NTP weigh’s equal to the molecular mass and contains Avogadro’s number

  $6.023\times {10}^{23}$ of particles

  $\begin{array}{l}no.ofmolesof=\frac{givenmass}{molarmass}\\ =\frac{5.53g}{44.1g/mol}\\ =0.125moles\end{array}$

  ${\mathrm{C}}_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

According to stoichiometry I mole of propane requires 5 moles of Oxygen. Requires

= 0.125 $\times$ 5 = 0.625 moles of oxygen.

According to the Ideal gas equation

  $\mathrm{PV}=nRT$

  $\mathrm{V}=\frac{nRT}{P}$

From this volume of a gas is calculated.

The equation is given by:

  ${\mathrm{Mg}}_3{N}_2(s)+3H_{2}O(l)\to 3MgO(s)+2N{H}_3(g)$

Molar mass of ${\mathrm{Mg}}_3{N}_2$ is calculated as:

  $(24.305\times 3)+(14.0067\times 2)=100.9284g/mol$

So. The number of moles of ${\mathrm{Mg}}_3{N}_2$ present in

no. of moles $\text{=}\frac{\text{mass}}{\text{molar mass}}$

  $\begin{array}{l}=\frac{10.3g}{100.9284g/mol}\\ =0.1021mol\end{array}$

From the above equation one mole of ${\mathrm{Mg}}_3{N}_2$ will produce 2 moles of ammonia on reaction with water

Hence, the number of moles of ammonia $(N{H}_3)$ evolved.

  $\begin{array}{l}=0.1021\times 2mol\\ =0.2042mol\end{array}$

Volume of ammonia =?

  $\mathrm{PV}=nRT$

  $\mathrm{V}=\frac{nRT}{P}$

  $\begin{array}{l}P=\frac{752}{760}=0.989\\ T=24+273=299\\ V=\frac{0.2042\times 0.0821\times 297}{0.989}\\ V=\frac{4.9791}{0.989}=5.034L\end{array}$

Volume of ammonia is $5.034L$

Conclusion

Volume of ammonia gas $5.034L$