Chapter 13, Problem 44A

Interpretation Introduction

Interpretation: -Volume of the chlorine gas needs to be calculated using the Ideal gas equation.

Concept Introduction: - For calculating the volume of chlorine, we use Ideal gas equation

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

Firstly, we calculate the no. of moles and then put values in the ideal gas equation for calculating volume.

Answer to Problem 44A

Volume of chlorine gas is $2.33ml$ .

Explanation of Solution

There is a relation between volume of a gas and number of molesand this relation is expressed by ideal gas equation

  $\mathrm{PV}=nRT$

For calculating volume, the formula is

  $\mathrm{V}=\frac{nRT}{P}$

under standard condition value of

  $\begin{array}{l}\text{T = 273K}\\ \text{P = 1 atm}\\ \text{R = 0}\text{.0821 = gas constant }\\ \text{n = no}\text{. of moles }\\ \text{V = volume }\end{array}$

No. of moles of $2Na=\frac{mass}{molarmass}$

  $\begin{array}{l}=\frac{4.18}{45.96}\\ =0.104moles.\end{array}$

According to Ideal gas equation,

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

Balanced equation is

  ${\mathrm{CaCO}}_3(s)\to CaO(s)+C{O}_2(g)$

Number of moles of ${\mathrm{CaCO}}_3$

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

  $\begin{array}{l}n=\frac{15.2g}{100.08g/mol}\\ n=0.150mol\end{array}$

According to Ideal gas equation,

  $\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

  $\begin{array}{l}P=1atm\\ V=?\\ T=273k\\ R=0.0821\\ n=numberofmoles=0.104\end{array}$

  $\mathrm{V}=\frac{nRT}{P}$

  $\begin{array}{l}\text{V = }\frac{\text{0}\text{.104 mol × 0}\text{.0821atm L × 273k}}{\text{1atm × k mol}}\\ \text{V = 0}\text{.104 × 0}\text{.0821L × 273}\\ \text{V = 2}\text{.33 L}\end{array}$

Volume of chlorine gas is $2.33ml$ .

Conclusion

Volume of chlorine gas is $2.33ml$ .