Question

Here is the question: picture is below for clarity. 

The zinc content of a 1.30$1.30$ g ore sample was determined by dissolving the ore in HCl$HCl$, which reacts with the zinc. The excess HCl$HCl$ is then neutralized with with NaOH$NaOH$. The reaction of HCl$HCl$ with Zn$Zn$ is shown.

Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)$$\text{Zn}(\text{s})+2\text{HCl}(\text{aq})⟶{\text{ZnCl}}_2^{}(\text{aq})+{\text{H}}_2^{}(\text{g})$$

The ore was dissolved in 150 mL$150\text{ mL}$ of 0.600 M HCl$0.600\text{ M }\text{HCl}$, and the resulting solution was diluted to a total volume of 300 mL$300\text{ mL}$. A 20.0 mL$20.0\text{ mL}$ aliquot of the final solution required 9.42$9.42$ mL of 0.510$0.510$ M NaOH$NaOH$ to neutralize the excess HCl$HCl$. What is the mass percentage (%w/w) of Zn$Zn$ in the ore sample?

Answer 1

Th reactions are,

$$ \begin{aligned} &\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2} \\ &\mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

Moles of \(\mathrm{HCl}\) is \(20 \mathrm{~mL}\) solution \(=\) moles of \(\mathrm{NaOH}\)

$$ \begin{aligned} =8.23 \times 0.510 \\ =4.20 \mathrm{mmol} \end{aligned} $$

moles of \(\mathrm{HCl}\) in \(300 \mathrm{~mL}\) solution \(=4.20 \mathrm{mmol} \times \frac{300}{20}\)

$$ =62.96 \mathrm{mmol} $$

Actual moles of \(\mathrm{HCl}=150 \times 0.6\)

$$ =90 \mathrm{mmol} $$

Reacted moles of \(\mathrm{HCl}=90-62.96\)

$$ =27.04 \mathrm{mmol} $$

1 mol \(Z\) n reacts with \(2 \mathrm{~mol} \mathrm{HCl}\).

Number of moles of \(\mathrm{HCl}=\frac{1}{2} \times 27.04 \mathrm{~mm} \mathrm{ol}\)

$$ \begin{aligned} &=13.52 \mathrm{mmol} \\ &=0.01352 \mathrm{~mol} \end{aligned} $$

$$ \text { Mass of } Z \mathrm{n}=0.01352 \mathrm{~mol} \times \frac{65.4 \mathrm{~g}}{1 \mathrm{~mol}} $$

$$ =0.88 \mathrm{~g} $$

Percent of \(Z \mathrm{n}=\frac{0.88 \mathrm{~g}}{1.30 \mathrm{~g}} \times 100\)

$$ =68.01 \% $$