Chapter 4, Problem 4.86PAE

Interpretation Introduction

Interpretation: The mass percentage of aluminum in the alloy should be found out if 0.3284 g of $H_2(g)$ was produced when a 7.264 g sample of a particular iron-aluminum alloy was dissolved in excess hydrochloric acid.

Concept introduction:

Iron- aluminum alloys possess a property of getting magnetized in the presence of a magnetic field and to get demagnetized in its absence.

When metals above Hydrogen in the reactivity series are reacted with hydrochloric acid, it will result in the metal chloride with bubbles of hydrogen gas.

Answer to Problem 4.86PAE

Solution:

Percentage of aluminum in alloy = 12.6 %

.

Explanation of Solution

Given information:

Total mass of Alloy = 7.264 g

Total mass of $H_2(g)$ released out = 0.3284 g

Aluminum and iron react with HCl as shown below.

$\begin{array}{l}2Al(s)+6HCl(aq)\to 2AlC{l}_3(aq)+ 3H_2(g)\\ Fe(s)+2HCl(aq)\to FeC{l}_2(aq)+H_2(g)\end{array}$

Assuming that,

Mass of Aluminum = m g

Mass of Iron = 7.264 - m g

Stoichiometry of,

Al: H₂ = 1: 3/2

Fe: H₂ = 1: 1

Because of excess HCl, it is assumed that total alloy reacted.

Therefore,

$\begin{array}{l}\mathrm{Re}acted\text{ number of moles of Al}= \raisebox{1ex}{$mg$}\!\left/ \!\raisebox{-1ex}{$27gmo{l}^{-1}$}\right.\hfill \\ Reacted\text{ number of moles of }Fe= \raisebox{1ex}{$7.264-m\text{ g}$}\!\left/ \!\raisebox{-1ex}{$56{\text{ gmol}}^{-1}$}\right.\hfill \\ \hfill \\ Evolved{H}_2(g) molsfromtotalAlmols =\left(m/27\right)\times 3/2mols\hfill \\ Evolved{H}_2(g) molsfromtotalFemols =\left[\left(7.264–m\right)/56\right]\times 1mols\hfill \\ \hfill \end{array}$

According to given data,

Evolved H₂(g) moles = 0.3284 g / 2 gmol-1

Stoichiometric $H_2(g)$ moles = practically evolved $H_2(g)$ moles

$\begin{array}{l}\text{(}m/27)\times 3/2mol+ \left[\left(7.264–m\right)/56\right]\times 1mol= 0.3284g/2gmo{l}^{-1}\\ \left(3m/54\right)mol+ \left[\left(7.264–m\right)/56\right]mol\text{ = 0}\text{.1642 mol}\\ \frac{168m+(392.256-54m)}{54\times 56}\text{ mol = 0}\text{.1642 mol}\\ \text{ 114 m = [0}\text{.1642}\times \text{(54}\times \text{56)]- 392}\text{.256}\\ \\ \text{ m =}\frac{104.2848}{114}\\ \text{ m = 0}\text{.915 g}\\ \\ \\ \end{array}$

$\begin{array}{l}Percentageofaluminum = \frac{0.915\text{ g}}{7.264\text{ g}}\times 100\%\\ \\ \text{ = 12}\text{.6\%}\end{array}$

Conclusion

Mass percentage of a metal in an alloy can be found out when the total mass of alloy and the mass of hydrogen gas released out when the alloy is reacted with hydrochloric acid is known.