Chapter 5, Problem 5.153QP

A sample of sodium peroxide, Na₂O₂, was reacted with an excess of water.

$2{\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}(s)+2{\text{H}}_{\text{2}}\text{O(}I\text{)}\to \text{4NaOH(}aq{\text{)+O}}_{\text{2}}(g)$

All of the sodium peroxide reacted, and the oxygen was collected over water at 21°C. The barometric pressure was 771 mmHg. The apparatus was similar to that shown in Figure 5.20. However, the level of water inside the tube was 25.0 cm above the level of water outside the tube. If the volume of gas in the tube is 31.0 mL, how many grams of sodium peroxide were in the sample?

Interpretation Introduction

Interpretation:

The amount of sodium peroxide (in grams) present in the sample after the reaction should be calculated.

${\text{2Na}}_{\text{2}}{\text{O}}_{\text{2}}{\text{(s)+2H}}_{\text{2}}\text{O(l)}\to {\text{4NaOH(aq)+O}}_{\text{2}}\text{(g)}$

Concept Introduction:

1. The relation between pressure and height of the barometer is given as

$\begin{array}{l}\text{ P}\text{=}\text{gdh}\\ \text{where ,}\\ \text{g =centre of gravity}\\ \text{d = density}\\ \text{h =height of the instrument}\\ \end{array}$

2. Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{nRT = PV}\\ \text{n = }\frac{\text{PV}}{\text{RT}}\end{array}$

3. $\text{No of moles = mass of the substance×}\frac{\text{1}\text{.0 mol}}{\text{molar mass of the substance}}$

Answer to Problem 5.153QP

The amount of sodium peroxide (in grams) sample present is $\text{0}\text{.194 grams}$.

Explanation of Solution

Given,

A sample of sodium peroxide with pressure $\text{771}\text{mmHg and }{21}^{\text{0}}\text{C }$.

The level of water inside the tube is $\text{25}\text{.0 cm}$ which is above level of water inside the tube.

The volume of gas in the tube = $\text{31}\text{.0}\text{ml}$.

${\text{2Na}}_{\text{2}}{\text{O}}_{\text{2}}{\text{(s)+2H}}_{\text{2}}\text{O(l)}\to {\text{4NaOH(aq)+O}}_{\text{2}}\text{(g)}$

Since level of water inside the tube is above the level of water outside the box

So subtract the level of water inside the tube

$\text{P}\text{=}\text{gdh}$

$\begin{array}{l}{\text{P}}_{\text{1}}\text{=}{\text{gd}}_{\text{Hg}}{\text{h}}_{\text{Hg}}\\ {\text{P}}_{\text{2}}{\text{=gd}}_{\text{W}}{\text{h}}_{\text{W}}\end{array}$

Compare P₁ and P₂ we will get

$\begin{array}{l}{\text{gd}}_{\text{W}}{\text{h}}_{\text{W}}\text{=}{\text{gd}}_{\text{Hg}}{\text{h}}_{\text{Hg}}\\ {\text{ h}}_{\text{Hg}}\text{=}\frac{{\text{gd}}_{\text{W}}{\text{h}}_{\text{W}}}{{\text{gd}}_{\text{Hg}}}\\ \text{=}\frac{\text{250mm×0}{\text{.99987 g/m}}^{\text{3}}}{\text{13}{\text{.596 g/m}}^{\text{3}}}\text{=18}\text{.39mmHg}\end{array}$

Pressure of the gas can be calculated as

$\begin{array}{l}{\text{P}}_{\text{gas}}{\text{=P-P}}_{\text{25}\text{cm}\text{water}}\\ \text{=771}\text{mmHg-18}\text{.39}\text{mmHg}\\ \text{=752}\text{.61mmHg}\\ \text{pressure}\text{of}{\text{O}}_{\text{2}}\text{gas}\text{calculated}\text{as}\\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}{\text{P}}_{\text{gas}}{\text{-P}}_{\text{water}}\\ \text{=}\text{752}\text{.61}\text{mmHg-18}\text{.7}\text{mmHg}\\ \text{=}\text{733}\text{.91mmHg}\\ \text{no}\text{of}\text{moles}\text{of}\text{sodium}\text{perioxide}\text{can}\text{be}\text{calculated}\text{as}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\left(\text{733}\text{.91/760atm}\right)\left(\text{0}\text{.0310L}\right)}{\left(\text{0}\text{.08206L}\text{.atm/K}\text{.mol}\right)\left(\text{294K}\right)}\\ \text{=}\text{0}\text{.001241}\text{mol}{\text{O}}_{\text{2}}\\ \text{Mass}\text{can}\text{be}\text{calculated}\text{as}\\ \text{Mass}\text{=}\text{0}{\text{.001241molO}}_{\text{2}}\text{×}\frac{\text{2}\text{mol}{\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}\text{×}\frac{\text{77}{\text{.98gNa}}_{\text{2}}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{O}}_{\text{2}}}\\ \text{=}\text{0}\text{.1935g}\end{array}$

Conclusion

The amount of sodium peroxide (in grams) present in the sample after the reaction is calculated as $\text{0}\text{.194 grams}$.