Q4.57P - Chemistry: Molecular Nature Of Matter And Change

Question

A mixture of bases can sometimes be the active ingredient in antacid tablets. If 0.4826 g of a mixture of ${\mathbf{\text{Al(OH)}}}_{\mathbf{\text{3}}}$ and  $\mathbf{\text{Mg}}{\mathbf{\left(}\mathbf{\text{OH}}\mathbf{\right)}}_{\mathbf{\text{2}}}$is neutralized with 17.30 mL of 1.000 M  ${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}\mathbf{\text{,}}$ what is the mass % of  ${\mathbf{\text{Al(OH)}}}_{\mathbf{\text{3}}}$ in the mixture?

Step-by-Step Solution

Answer

The mass % of  $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ in the mixture is 55.93%

Step 1: Write down the molecular equation

Balanced molecular equation:

$\mathrm{Al}{\left(\mathrm{OH}\right)}_3+\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+5{\mathrm{HNO}}_3\to \mathrm{Al}{\left({\mathrm{NO}}_3\right)}_3+\mathrm{Mg}{\left({\mathrm{NO}}_3\right)}_2+5{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2: Calculate the mass of Al(OH) 3

$\begin{array}{c}{\text{number\hspace{0.17em}of\hspace{0.17em}moles\hspace{0.17em}of\hspace{0.17em}HNO}}_{\text{3}}=1\text{\hspace{0.17em}mol/l×17.3ml×}\frac{\text{1l}}{\text{1000ml}}\\ \text{=0.0173 mole}\end{array}$

$\text{mole\hspace{0.17em}\hspace{0.17em}ratio\hspace{0.17em}=\hspace{0.17em}}\frac{1\text{\hspace{0.17em}\hspace{0.17em}mole Al}{\left(\text{OH}\right)}_3}{5{\text{\hspace{0.17em}moles HNO}}_3}=\frac{1}{5}$

Number of moles of  $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$  that reacted with  $0.0173\text{\hspace{0.17em}moles}$ of ${\mathrm{HNO}}_3$ is calculated below.

$\begin{array}{c}\text{Number\hspace{0.17em}of\hspace{0.17em}moles of\hspace{0.17em}Al}{\left(\text{OH}\right)}_{\text{3}}=0.0173\times \frac{1}{5}\\ =0.00346\end{array}$

$\begin{array}{c}\mathrm{Mass}\mathrm{of}1\mathrm{mole}\mathrm{Al}{\left(\mathrm{OH}\right)}_3=78.01\mathrm{g}\\ \mathrm{Mass}\mathrm{of}0.00346\mathrm{mole}\mathrm{Al}{\left(\mathrm{OH}\right)}_3=78.01\mathrm{g}\times 0.00346\\ =0.2699\mathrm{g}\end{array}$

Step 3: Determine the mass % of Al(OH) 3

The mass percent of ${\mathbf{\text{Al(OH)}}}_{\mathbf{\text{3}}}\mathbf{:}$

$\begin{array}{ccc}\mathrm{mass}\mathrm{\%} \mathrm{of} \mathrm{Al}{\left(\mathrm{OH}\right)}_3& =& \frac{\mathrm{mass} \mathrm{of} \mathrm{Al}{\left(\mathrm{OH}\right)}_3}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{mixture} \mathrm{of} \mathrm{Al}{\left(\mathrm{OH}\right)}_3 \mathrm{and} \mathrm{Mg}{\left(\mathrm{OH}\right)}_2}\times 100\mathrm{\%}\\ & =& \frac{0.2699 \mathrm{gmAl}{\left(\mathrm{OH}\right)}_3}{0.4826\mathrm{gm}}\mathrm{x}100\mathrm{\%}\\ & =& 55.93\mathrm{\%}\end{array}$