Commercial brass, an alloy of \(Z n\) and \(\mathrm{Cu}\), reacts with hydrochloric acid as follows: $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ (Cu does not react with HCl.) When \(0.5065 \mathrm{~g}\) of a certain brass alloy is reacted with excess \(\mathrm{HCl}, 0.0985 \mathrm{~g} \mathrm{ZnCl}_{2}\) is eventually isolated. a. What is the composition of the brass by mass? b. How could this result be checked without changing the above procedure?

Given that the alloy weight of 0.5065 g reacts with excess HCl, we're told that 0.0985 g of ZnCl2 is formed. First, we need to find how much zinc is in the isolated 0.0985 g of ZnCl2. The molar mass of Zn is 65.38 g/mol, and the molar mass for ZnCl2 is 136.30 g/mol (65.38 g/mol + (2 * 35.45 g/mol)). Now, we'll find the amount of Zn in moles within ZnCl2: $$ \text{moles of Zn} = \frac{\text{mass of ZnCl}_2}{\text{molar mass of ZnCl}_2} = \frac{0.0985 \mathrm{~g}}{136.30 \mathrm{~g/mol}} = 0.000722 \mathrm{~mol} $$ Next, we'll find the mass of Zn in the alloy: $$ \text{mass of Zn} = \text{moles of Zn} \times \text{molar mass of Zn} = 0.000722 \mathrm{~mol} \times 65.38 \mathrm{~g/mol} = 0.0472 \mathrm{~g} $$ The mass of Zn in the alloy is 0.0472 g.

Since the alloy consists of only Zn and Cu, we can find the mass of Cu in the sample by subtracting the mass of Zn from the total mass of the alloy. $$ \text{mass of Cu} = \text{mass of alloy} - \text{mass of Zn} = 0.5065 \mathrm{~g} - 0.0472 \mathrm{~g} = 0.4593 \mathrm{~g} $$ The mass of Cu in the alloy is 0.4593 g.

Now, we'll calculate the mass percentages of Zn and Cu in the alloy. For Zn: $$ \text{percentage of Zn} = \frac{\text{mass of Zn}}{\text{mass of alloy}} \times 100\% = \frac{0.0472 \mathrm{~g}}{0.5065 \mathrm{~g}} \times 100\% = 9.32\% $$ For Cu: $$ \text{percentage of Cu} = \frac{\text{mass of Cu}}{\text{mass of alloy}} \times 100\% = \frac{0.4593 \mathrm{~g}}{0.5065 \mathrm{~g}} \times 100\% = 90.68\% $$ The composition of the brass alloy by mass is around 9.32% Zn and 90.68% Cu.

To answer part b and check the result without changing the procedure, we could use the following method: Weigh the sample again once the reaction with HCl is complete, ensuring that the Cu and any excess HCl are removed completely. The residual mass should be close to the mass of Cu that was calculated in Step 2, which is 0.4593 g.