Chapter 5, Problem 5.100QP

(a)

Interpretation Introduction

Interpretation:

The given reaction of equation and the percent composition by mass of the mixture, the volume of the solids has to be determined.

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 5.100QP

$\begin{array}{l}{\text{CaO}}_{\text{(s)}}{\text{+CO}}_{\text{2(g)}}\to {\text{CaCO}}_{\text{3(s)}}\\ {\text{BaO}}_{\text{(s)}}{\text{+CO}}_{\text{2(g)}}\to {\text{BaCO}}_{\text{3(s)}}\end{array}$

Explanation of Solution

To find the given reaction of equation

$\begin{array}{l}{\text{CaO}}_{\text{(s)}}{\text{+CO}}_{\text{2(g)}}\to {\text{CaCO}}_{\text{3(s)}}\\ {\text{BaO}}_{\text{(s)}}{\text{+CO}}_{\text{2(g)}}\to {\text{BaCO}}_{\text{3(s)}}\end{array}$

Calcium oxide is combined with Carbon dioxide to give Calcium Carbonate this reaction formed addition product.

Barium oxide is combined with Carbon dioxide to give Barium Carbonate this reaction formed addition product.

(b)

Interpretation Introduction

Interpretation:

The given reaction of equation and the percent composition by mass of the mixture, the volume of the solids has to be determined.

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 5.100QP

$\begin{array}{l}\text{CaO : }\frac{\text{0}\text{.513 g}}{\text{4}\text{.88 g}}\times \text{100\%}\text{=}\text{10}\text{.5\%}\\ \text{BaO : }\frac{\text{4}\text{.37 g}}{\text{4}\text{.88 g}}\times \text{100\%}\text{=}89.\text{5\%}\end{array}$

Explanation of Solution

We need to find the number of moles of ${\text{CO}}_{\text{2}}$ consumed in the reaction. We can do this by calculating the initial moles of ${\text{CO}}_{\text{2}}$ in the flask and then comparing it to the ${\text{CO}}_{\text{2}}$ remaining after the reaction.

$\begin{array}{l}\text{initially:}\\ {}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\left(\text{746}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)\text{(1}\text{.46}\text{L)}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(35+273}\text{K)}}\text{=}\text{0}\text{.0567}\text{mol}{\text{CO}}_{\text{2}}\\ \text{remaining:}\\ {\text{n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\left(\text{746}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)\text{(1}\text{.46}\text{L)}}{\left(\text{0}\text{.08206}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(35}\text{+273}\text{K)}}\text{=}\text{0}\text{.0191}\text{mol}{\text{CO}}_{\text{2}}\end{array}$

$\text{mol CaO + mol BaO = 0}\text{.0376 mol }$

Thus, the amount of ${\text{CO}}_{\text{2}}$ consumed in the reaction is:

$\left(\text{0}\text{.0567 mol - 0}\text{.0191 mol}\right)\text{ = 0}\text{.0376 mol}$ ${\text{CO}}_{\text{2}}$. Since the mole ratio between ${\text{CO}}_{\text{2}}$ and both reactants ($\text{CaO and BaO) is 1:1, 0}\text{.0376 mole}$) of the mixture must have reacted.

$\text{Let x = mass of CaO in the mixture, then (4}\text{.88 - x) = mass of BaO in the mixture}\text{. We can write: }$

$\left[\text{x}\text{g}\text{CaO}\text{×}\frac{\text{1mol}\text{CaO}}{\text{56}\text{.08}\text{g}\text{CaO}}\right]\text{+}\left[\text{(4}\text{.88-x)g}\text{BaO}\text{×}\frac{\text{1}\text{mol}\text{BaO}}{\text{153}\text{.3g}\text{BaO}}\right]$

$\begin{array}{l}\text{0}\text{.01783x - 0}\text{.006523x + 0}\text{.0318 = 0}\text{.0376 }\\ \\ \text{x = 0}\text{.513 g = mass of CaO in the mixture }\\ \\ \text{mass of BaO in the mixture = 4}\text{.88 - x = 4}\text{.37 g }\end{array}$

Mass of Calcium oxide in the mixture $\text{0}\text{.513 g }$ so the

$\text{mass of BaO in the mixture = 4}\text{.88 - x = 4}\text{.37 g }$

$\begin{array}{l}\text{The percent compositions by mass in the mixture are: }\\ \\ \text{CaO : }\frac{\text{0}\text{.513 g}}{\text{4}\text{.88 g}}\times \text{100\%}\text{=}\text{10}\text{.5\%}\\ \text{BaO : }\frac{\text{4}\text{.37 g}}{\text{4}\text{.88 g}}\times \text{100\%}\text{=}89.\text{5\%}\end{array}$

Hence the mixture is contains $\text{10}\text{.5\%}$ of the Calcium oxide and $89.\text{5\%}$ of Barium oxide