Chapter 5, Problem 5.91PAE

91 A 0.2500-g sample of an Al-Zn alloy reacts with HCl to form hydrogen gas:

   $\text{2Al}\left(\text{s}\right)+{\text{6H}}^{+}\left(\text{aq}\right)\to {\text{2Al}}^{\text{3}+}\left(\text{aq}\right)+{\text{3H}}_{\text{2}}\left(\text{g}\right)$

   $\text{Zn}\left(\text{s}\right)+{\text{2H}}^{+}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2}+}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

The hydrogen produced has a volume of 0.147 L at 25°C and 755 mm Hg. What is the percentage of zinc in the alloy?

Interpretation Introduction

Interpretation:

A 0.2500 g sample of an Al-Zn alloy reacts with HCl to form hydrogen gas:

$\begin{array}{l}\text{2Al}\left(\text{s}\right)+{\text{6H}}^{+}\left(\text{aq}\right)\to {\text{2Al}}^{\text{3}+}\left(\text{aq}\right)+{\text{3H}}_{\text{2}}\left(\text{g}\right)\\ \text{Zn }\left(\text{s}\right)+{\text{2 H}}^{+}\left(\text{aq}\right)\to {\text{Zn}}^{\text{2}+}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\end{array}$

The hydrogen produced has a volume of 0.147L at 25⁰C and 755mm Hg. The percentage of Zinc in the alloy needs to be calculated.

Concept introduction:

An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the ideal gas laws: -

$\text{PV = nRT}$

Here,

n = Number of moles of the gas

R = Universal gas constant

T = Temperature

P = Pressure

Ideal gases are described as the molecules which have negligible size but have an average molar kinetic energy which is dependent on the temperature. When temperature is low most of the gases behave like ideal gases and the ideal gas law might be applied to them.

Answer to Problem 5.91PAE

Solution:

$78.7\%$

Explanation of Solution

Here given that,

Temperature = 25⁰C;

Converting Temperature into Kelvin

$\begin{array}{l}{\text{T (K) = 273 + T(}}^0\text{C)}\\ {\text{ = 273+25}}^0\text{C}\\ \text{ = 298 K}\end{array}$

Therefore, the temperature in Kelvin is 298 K

Pressure = 755 mm Hg = Converting mmHg to atm

$\frac{1}{760}\times 755=0.993\text{atm}$

Therefore, the Pressure in atm is 0.993 atm

The number of moles gas is calculated from ideal gas equation.

${\text{P}}_{gas}\text{ =}\frac{{\text{n}}_{gas}\text{RT}}{\text{V}}$

Total number hydrogen moles can be calculated by using an ideal gas equation: -

${\text{P}}_{H_2}\text{ =}\frac{{\text{n}}_{H_2}\text{RT}}{\text{V}}$

Here ${\text{P}}_{H_2}$ = Pressure of gas = 0.993 atm

${\text{n}}_{H_2}$ = number of moles =

R = Gas constant = 0.0821 L.atm / mol. K

T = Temperature = 298

V = Volume = 0.147 L

$\begin{array}{l}0.993\text{ =}\frac{{\text{n}}_{{\text{H}}_2}\times \frac{0.821\text{ L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\times 298K}{\text{0}\text{.147 L}}\\ {\text{n}}_{{\text{H}}_2}=\left(\frac{0.993\times \text{0}\text{.147}}{0.821\times 298}\right)\text{mol}\\ \text{ = 0}\text{.00597 mol}\end{array}$

Total number of hydrogen moles comes by the combination of Al and Zn

Now, let’s assume mass of Al is x therefore, mass of Zn is (0.2500-x) g

Therefore,

$\begin{array}{c}\text{0}{\text{.00597 mol H}}_2\text{ = }\left[xg\text{ of Al }\times \frac{1\text{ mol of Al}}{26.98\text{g of Al}}\times \frac{{\text{3/2 mol of H}}_2}{1\text{ mol of Al}}\right]\\ \times \left[\left(0.2500-x\right)\text{g of Zn}\times \frac{1\text{ mol of Zn}}{65.39\text{g of Zn}}\times \frac{1{\text{ mol of H}}_2}{1\text{mol of Zn}}\right]\\ \text{0}\text{.00597 = 0}\text{.00556}x\text{ + 0}\text{.00382}-\text{ 0}\text{.0153}x\\ 0.00597-0.00382=\left(\text{0}\text{.00556}-\text{0}\text{.0153}\right)x\\ x=\text{ 0}\text{.0533 g of Al}\end{array}$

Now the mass of Zn will be: -

$\begin{array}{c}{\text{m}}_{\text{Zn}}=\left(0.2500-x\right)\text{ g}\\ \text{=}\left(\text{0}\text{.2500}-0.5333\right)\text{ g}\\ \text{= 0}\text{.197 g of Zn }\end{array}$

Now, calculating the percentage of Zn by using below equation: -

$\text{Zn\% = }\frac{{\text{mass}}_{\text{Zn}}}{\text{mass of alloy}}\times 100\%$

Now substituting the value of Zn and alloy in the above equation: -

$\begin{array}{l}\text{Zn\% = }\frac{{\text{mass}}_{\text{Zn}}}{\text{mass of alloy}}\times 100\%\\ \text{ = }\frac{0.197\text{ g}}{0.2500\text{ g}}\times 100\%\\ \text{ = 78}\text{.7\% }\end{array}$

Conclusion

An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the ideal gas laws: -

$\text{PV = nRT}$

Here,

n = Number of moles of the gas

R = Universal gas constant

T = Temperature

P = Pressure

The above equation can be used to calculate the percentage of Zn in the alloy.