Chapter 5, Problem 5.154QP

Interpretation Introduction

Interpretation:

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

$\text{Zn(s)+2HCl(aq)}\to {\text{ZnCl}}_{\text{2}}{\text{(aq)+H}}_{\text{2}}\text{(g)}$

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 5.154QP

The amount of Zinc (in grams) sample present is $\text{0}\text{.0574 grams}$.

Explanation of Solution

Given,

A sample of Zinc with pressure ${\text{751 mmHg and temperature 17}}^{\text{0}}\text{C}$.

The level of water inside the tube is $\text{29}\text{.7 cm}$ which is above level of water inside the tube.

The volume of gas in the tube = $\text{22}\text{.10 mL}$

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

$\begin{array}{l}{\text{gd}}_{\text{W}}{\text{h}}_{\text{W}}\text{=}{\text{gd}}_{\text{Hg}}{\text{h}}_{\text{Hg}}\\ {\text{ h}}_{\text{Hg}}\text{=}\frac{{\text{gd}}_{\text{W}}{\text{h}}_{\text{W}}}{{\text{gd}}_{\text{Hg}}}\\ \text{=}\frac{\text{297mm×0}{\text{.99987 g/m}}^{\text{3}}}{\text{13}{\text{.596 g/m}}^{\text{3}}}\text{=21}\text{.84mmHg}\end{array}$

Pressure of the gas can be calculated as

$\begin{array}{l}{\text{P}}_{\text{gas}}{\text{=P-P}}_{\text{31}\text{cm}\text{water}}\\ \text{=751}\text{mmHg-21}\text{.84}\text{mmHg}\\ \text{=729}\text{.21mmHg}\\ {\text{vapour pressure of water at 17}}^{\text{0}}\text{C}\\ {\text{P}}_{{\text{H}}_{\text{2}}}\text{=}{\text{P}}_{\text{gas}}{\text{-P}}_{\text{water}}\\ \text{=}\text{752}\text{.61}\text{mmHg-14}\text{.5}\text{mmHg}\\ \text{=}\text{714}\text{.7mmHg}\\ \text{no}\text{of}\text{moles}\text{of}\text{sodium}\text{perioxide}\text{can}\text{be}\text{calculated}\text{as}\\ \text{n}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\left(\text{714}\text{.7/760atm}\right)\left(\text{0}\text{.0221L}\right)}{\left(\text{0}\text{.08206L}\text{.atm/K}\text{.mol}\right)\left(\text{290K}\right)}\\ \text{=}\text{0}\text{.0008733}\text{mol}{\text{O}}_{\text{2}}\\ \text{Mass}\text{can}\text{be}\text{calculated}\text{as}\\ \text{Mass}\text{=}\text{0}\text{.0008733}\text{mol}{\text{O}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}\text{Zn}}{\text{1}\text{mol}{\text{H}}_{\text{2}}}\text{×}\frac{\text{65}{\text{.38gNa}}_{\text{2}}{\text{O}}_{\text{2}}}{\text{1}\text{mol}\text{Zn}}\\ \text{=}\text{0}\text{.057098 g}\end{array}$

Conclusion

The amount of Zinc (in grams) present in the sample after the reaction was calculated as $\text{0}\text{.0574 grams}$