Chapter 4, Problem 4.137P

(a)

Interpretation Introduction

Interpretation:

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy that has a mass of $0.263\text{ g}$ is to be determined.

Concept introduction:

Mass percent is employed to determine the concentration of one compound in a mixture of the compound. The formula to calculate mass percent is as follows:

$\text{Mass percent of compound}=\left(\frac{\text{mass of the compound}}{\text{mass of mixture}}\right)$

Answer to Problem 4.137P

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy is $22.7\%$.

Explanation of Solution

Consider the mass of $\text{Mg}$ is $x\text{g}$ and mass of $\text{Al}$ is $\left(0.263-x\right)\text{g}$.

The formula to calculate the volume of the alloy is as follows:

$\text{Volume of alloy}=\left(\frac{\text{mass of alloy}}{\text{density of alloy}}\right)$ (2)

Consider $0.263\text{ g}$ the mass of alloy and $2.40{\text{g/cm}}^{\text{3}}$ for the density of alloy in the equation (2).

$\begin{array}{c}\text{Volume of alloy}=\left(\frac{0.263\text{ g}}{2.40{\text{g/cm}}^{\text{3}}}\right)\\ =0.10958{\text{cm}}^{\text{3}}\end{array}$

The formula to calculate the volume of $\text{Mg}$ is as follows:

$\text{Volume of Mg}=\left(\frac{\text{mass of Mg}}{\text{density of Mg}}\right)$ (3)

Consider $x\text{g}$ the mass of $\text{Mg}$ and $1.74{\text{g/cm}}^{\text{3}}$ for the density of $\text{Mg}$ in the equation (3).

$\text{Volume of Mg}=\left(\frac{x\text{g}}{1.74{\text{g/cm}}^{\text{3}}}\right)$

The formula to calculate the volume of $\text{Al}$ is as follows:

$\text{Volume of Al}=\left(\frac{\text{mass of Al}}{\text{density of Al}}\right)$ (4)

Consider $\left(0.263-x\right)\text{g}$ the mass of $\text{Al}$ and $2.70{\text{g/cm}}^{\text{3}}$ for the density of $\text{Al}$ in the equation (4).

$\text{Volume of Al}=\left(\frac{\left(0.263-x\right)\text{g}}{2.70{\text{g/cm}}^{\text{3}}}\right)$

The formula to calculate $x$ is as follows:

$\text{Volume of alloy}=\text{Volume of Mg}+\text{Volume of Al}$ (5)

Substitute $0.10958{\text{cm}}^{\text{3}}$ for the volume of alloy, $\left(\frac{x\text{g}}{1.74{\text{g/cm}}^{\text{3}}}\right)$ for the volume of $\text{Mg}$ and $\left(\frac{\left(0.263-x\right)\text{g}}{2.70{\text{g/cm}}^{\text{3}}}\right)$ for the volume of $\text{Al}$ in the equation (5).

$\begin{array}{c}0.10958{\text{cm}}^{\text{3}}=\left(\frac{x\text{g}}{1.74{\text{g/cm}}^{\text{3}}}\right)+\left(\frac{\left(0.263-x\right)\text{g}}{2.70{\text{g/cm}}^{\text{3}}}\right)\\ x=0.05957\text{g}\end{array}$

The expression to calculate the mass percent of $\text{Mg}$ is:

$\text{mass }\%\text{of}\text{Mg}=\left(\frac{\text{mass}\text{of Mg}\left(\text{g}\right)}{\text{mass of alloy sample}\left(\text{g}\right)}\right)\left(100\right)$ (6)

Substitute $0.05957\text{g}$ for the mass of $\text{Mg}$ and $0.263\text{ g}$ for the mass of alloy sample in the equation (6).

$\begin{array}{c}\text{Mass }\%\text{of}\text{Mg}=\left(\frac{0.05957\text{g}}{0.263\text{ g}}\right)\left(100\right)\\ =22.6502\%\\ \approx 22.7\%\end{array}$

Conclusion

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy is $22.7\%$.

(b)

Interpretation Introduction

Interpretation:

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy that reacts with excess aqueous $\text{HCl}$ and forms $1.38\times {10}^{-2}\text{mol}{\text{H}}_2$ is to be determined.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 4.137P

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy is $21.6\%$.

Explanation of Solution

The reaction of $\text{Mg}$ and $\text{Al}$ with $\text{HCl}$ is as follows:

$\begin{array}{l}\text{Mg}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{MgCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)\\ 2\text{Al}\left(s\right)+6\text{HCl}\left(aq\right)\to 2{\text{AlCl}}_3\left(aq\right)+3{\text{H}}_{\text{2}}\left(\text{g}\right)\end{array}$

Consider the mass of $\text{Mg}$ is $x\text{g}$ and mass of $\text{Al}$ is $\left(0.263-x\right)\text{g}$.

The formula to calculate moles of ${\text{H}}_2$ from $\text{Mg}$ is as follows:

${\text{moles of H}}_2=\left(\frac{\text{mass of Mg}}{\text{molar mass of Mg}}\right)\left(\frac{\text{1}{\text{mol H}}_2}{1\text{mol Mg}}\right)$ (7)

Substitute $x\text{g}$ for the mass of $\text{Mg}$ and $24.31\text{g/mol}$ for molar mass of $\text{Mg}$ in the equation (7).

${\text{moles of H}}_2=\left(\frac{x\text{g}}{24.31\text{g}}\right)\left(\frac{\text{1}{\text{mol H}}_2}{1\text{mol Mg}}\right)$

The formula to calculate moles of ${\text{H}}_2$ from $\text{Al}$ is as follows:

${\text{moles of H}}_2=\left(\frac{\text{mass of Al}}{\text{molar mass of Al}}\right)\left(\frac{\text{3}{\text{mol H}}_2}{2\text{mol Al}}\right)$ (8)

Substitute $\left(0.263-x\right)\text{g}$ for the mass of $\text{Al}$ and $26.98\text{g/mol}$ for molar mass of $\text{Al}$ in the equation (8)

${\text{moles of H}}_2=\left(\frac{\left(0.263-x\right)\text{g}}{26.98\text{g}}\right)\left(\frac{\text{3}{\text{mol H}}_2}{2\text{mol Al}}\right)$

The formula to calculate $x$ is as follows:

${\text{moles of H}}_{\text{2}}\text{ produced}={\text{moles of H}}_{\text{2}}\text{ from Mg}+{\text{moles of H}}_{\text{2}}\text{from Al}$ (9)

Substitute $\left(\frac{x\text{g}}{24.31\text{g/mol}}\right)\left(\frac{\text{1}{\text{mol H}}_2}{1\text{mol Mg}}\right)$ for moles of ${\text{H}}_2$ from $\text{Mg}$, $1.38\times {10}^{-2}\text{mol}$ for moles of ${\text{H}}_2$ produced and $\left(\frac{\left(0.263-x\right)\text{g}}{26.98\text{g/mol}}\right)\left(\frac{\text{3}{\text{mol H}}_2}{2\text{mol Al}}\right)$ for moles of ${\text{H}}_2$ from $\text{Al}$ in the equation (9).

$\begin{array}{c}1.38\times {10}^{-2}\text{mol}=\left(\frac{x\text{g}}{24.31\text{g/mol}}\right)\left(\frac{\text{1}{\text{mol H}}_2}{1\text{mol Mg}}\right)+\left(\frac{\left(0.263-x\right)\text{g}}{26.98\text{g/mol}}\right)\left(\frac{\text{3}{\text{mol H}}_2}{2\text{mol Al}}\right)\\ 8.22\times {10}^{-4}=0.014462x\\ x=0.05684\text{g}\end{array}$

Substitute $0.05684\text{g}$ for the mass of $\text{Mg}$ and $0.263\text{ g}$ for the mass of alloy sample in the equation (6).

$\begin{array}{c}\text{Mass }\%\text{of}\text{Mg}=\left(\frac{0.05684\text{g}}{0.263\text{ g}}\right)\left(100\right)\\ =21.6122\%\\ \approx 21.6\%\end{array}$

Conclusion

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy is $21.6\%$.

(c)

Interpretation Introduction

Interpretation:

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy that reacts with excess ${\text{O}}_2$ and forms $0.483\text{g}$ of oxide is to be determined.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

$\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 4.137P

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy is $22.9\%$.

Explanation of Solution

The reaction of $\text{Mg}$ and $\text{Al}$ with ${\text{O}}_2$ is as follows:

$\begin{array}{l}\text{Mg}\left(s\right)+{\text{O}}_2\left(aq\right)\to 2\text{MgO}\left(s\right)\\ 4\text{Al}\left(s\right)+3{\text{O}}_2\left(aq\right)\to 2{\text{Al}}_2{\text{O}}_3\left(s\right)\end{array}$

Consider the mass of $\text{Mg}$ is $x\text{g}$ and mass of $\text{Al}$ is $\left(0.263-x\right)\text{g}$.

The formula to calculate the mass of $\text{MgO}$ from $\text{Mg}$ is as follows:

$\text{mass of MgO}=\left(\frac{\text{mass of Mg}}{\text{molar mass of Mg}}\right)\left(\frac{\text{2}\text{mol MgO}}{2\text{mol Mg}}\right)\left(\text{molar mass of MgO}\right)$ (10)

Substitute $x\text{g}$ for the mass of $\text{Mg}$, $24.31\text{g/mol}$ for molar mass of $\text{Mg}$ and $40.31\text{g/mol}$ for the molar mass of $\text{MgO}$ in the equation (10).

$\text{mass of MgO}=\left(\frac{x\text{g}}{24.31\text{g/mol}}\right)\left(\frac{\text{2}\text{mol MgO}}{2\text{mol Mg}}\right)\left(40.31\text{g/mol}\right)$

The formula to calculate the mass of ${\text{Al}}_2{\text{O}}_3$ from $\text{Al}$ is as follows:

${\text{mass of Al}}_2{\text{O}}_3=\left(\frac{\text{mass of Al}}{\text{molar mass of Al}}\right)\left(\frac{\text{2}{\text{mol Al}}_2{\text{O}}_3}{4\text{mol Al}}\right)\left({\text{molar mass of Al}}_2{\text{O}}_3\right)$ (11)

Substitute $\left(0.263-x\right)\text{g}$ for the mass of $\text{Al}$, $26.98\text{g/mol}$ for molar mass of $\text{Al}$ and $101.96\text{g/mol}$ for the molar mass of ${\text{Al}}_2{\text{O}}_3$ in the equation (11).

${\text{mass of Al}}_2{\text{O}}_3=\left(\frac{\left(0.263-x\right)\text{g}}{26.98\text{g/mol}}\right)\left(\frac{\text{2}{\text{mol Al}}_2{\text{O}}_3}{4\text{mol Al}}\right)\left(101.96\text{g/mol}\right)$

The formula to calculate $x$ is as follows:

$\text{mass of oxide produced}=\text{mass of MgO}\text{from Mg}+{\text{mass of Al}}_2{\text{O}}_3\text{from Al}$ (12)

Substitute $\left(\frac{x\text{g}}{24.31\text{g/mol}}\right)\left(\frac{\text{2}\text{mol MgO}}{2\text{mol Mg}}\right)\left(40.31\text{g/mol}\right)$ for the mass of $\text{MgO}$ from $\text{Mg}$, $0.483\text{g}$ for moles of oxide produced and $\left(\frac{\left(0.263-x\right)\text{g}}{26.98\text{g/mol}}\right)\left(\frac{\text{2}{\text{mol Al}}_2{\text{O}}_3}{4\text{mol Al}}\right)\left(101.96\text{g/mol}\right)$ for moles of ${\text{Al}}_2{\text{O}}_3$ from $\text{Al}$ in the equation (12).

$\begin{array}{c}0.483\text{g}=\left[\begin{array}{l}\left(\frac{x\text{g}}{24.31\text{g/mol}}\right)\left(\frac{\text{2}\text{mol MgO}}{2\text{mol Mg}}\right)\left(40.31\text{g/mol}\right)+\\ \left(\frac{\left(0.263-x\right)\text{g}}{26.98\text{g/mol}}\right)\left(\frac{\text{2}{\text{mol Al}}_2{\text{O}}_3}{4\text{mol Al}}\right)\left(101.96\text{g/mol}\right)\end{array}\right]\\ 0.01359=0.2315x\\ x=0.060298\text{g}\end{array}$

Substitute $0.060298\text{g}$ for the mass of $\text{Mg}$ and $0.263\text{ g}$ for the mass of alloy sample in the equation (6).

$\begin{array}{c}\text{Mass }\%\text{of}\text{Mg}=\left(\frac{0.060298\text{g}}{0.263\text{ g}}\right)\left(100\right)\\ =22.927\%\\ \approx 22.9\%\end{array}$

Conclusion

The mass percentage of $\text{Mg}$ in a magnesium-aluminium alloy is $22.9\%$.