Chapter 11, Problem 11.112QP

Interpretation Introduction

Interpretation:

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T.  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

$\text{PV = nRT}$

Answer to Problem 11.112QP

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture is $\text{33}\text{.1\%}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

Explanation of Solution

Calculate the moles of ${\text{CO}}_{\text{2}}$ produced using the ideal gas equation.  We carry an extra significant figure throughout this calculation to limit rounding errors.

$\begin{array}{l}{\text{ n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{(1}\text{.24}\text{atm)(1}\text{.67}\text{L)}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(26+}\text{273)}\text{K}}\text{=}\text{0}\text{.08436}\text{mol}{\text{CO}}_{\text{2}}\end{array}$

The moles of ${\text{CO}}_{\text{2}}$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.  The moles of ${\text{CO}}_{\text{2}}$ produced was found to be $\text{0}\text{.08436}\text{mol}{\text{CO}}_{\text{2}}$

Since there is a 1:1 mole ratio between ${\text{CO}}_{\text{2}}$ and both reactants $\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3 }}{\text{and MgCO}}_{\text{3}}\right)\text{,}$ then 0.08436 mole of the mixture must have reacted.

$\text{mol}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{+}\text{mol}{\text{MgCO}}_{\text{3}}\text{=}\text{0}\text{.08436}\text{mol}$

${\text{Let x be the mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ in the mixture, then (7}{\text{.63 - x) is the mass of MgCO}}_{\text{3 }}\text{in the mixture}\text{.}$ $\begin{array}{l}\left(\text{x}\text{g}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{106}\text{.0}\text{g}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\text{+}\left[\text{(7}\text{.63-x)g}{\text{MgCO}}_{\text{3}}\text{×}\frac{\text{1}{\text{molMgCO}}_{\text{3}}}{\text{84}\text{.32}\text{g}{\text{MgCO}}_{\text{3}}}\right]\\ \text{=}\text{0}\text{.08436}\text{mol}\end{array}$

$\text{x = 2}{\text{.527 g = mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ in the mixture}$

The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ mixture is calculated by plugging in the given weight of Magnesium sulphate, molecular weight of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and molecular weight of Magnesium sulphate.  The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ mixture was found to be $\text{x = 2}\text{.527 g}$

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$in the mixture is:

$\text{mass\%}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{×}\frac{\text{mass}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{mass}\text{of}\text{mixture}}\text{×}\text{100\%}\text{=}\text{33}\text{.1\%}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture is calculated using the ratio between $\text{mass}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ and mass of mixture.  The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture is $\text{33}\text{.1\%}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

Conclusion

The percent composition by mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ in the mixture was calculated.