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A 6.43-g sample of a $\mathrm{Cu}-\mathrm{Zn}$ alloy reacts with $\mathrm{HCl}$ acid to produce hydrogen gas. If the hydrogen gas has a volume of $1.33 \mathrm{~L}$ at $20^{\circ} \mathrm{C}$ and $763 \mathrm{mmHg}$, what is the percent of $\mathrm{Zn}$ in the alloy?
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∻Solution∻ ‖ 1 ⋮ **Convert the given volume of hydrogen gas to moles using the ideal gas law**: The ideal gas law is given by $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin. First, we need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature and the pressure to atmospheres by dividing by 760 mmHg/atm. ‖ ‖ 2 ⋮ **Calculate the moles of hydrogen gas produced**: Using the ideal gas law, $n = \frac{PV}{RT}$. We plug in the values: $P = \frac{763}{760} ext{ atm}$, $V = 1.33 ext{ L}$, $R = 0.0821 ext{ L atm mol}^{-1} ext{K}^{-1}$, and $T = 20 + 273.15 ext{ K}$. ‖ ‖ 3 ⋮ **Determine the moles of Zn that reacted**: Since the reaction between Zn and HCl produces H2 in a 1:1 molar ratio, the moles of Zn that reacted will be equal to the moles of H2 produced. ‖ ‖ 4 ⋮ **Calculate the mass of Zn that reacted**: The molar mass of Zn is approximately $65.38 ext{ g/mol}$. Multiply the moles of Zn by its molar mass to get the mass of Zn that reacted. ‖ ‖ 5 ⋮ **Determine the percentage of Zn in the alloy**: The percentage of Zn is calculated by dividing the mass of Zn that reacted by the total mass of the alloy sample and then multiplying by 100 to get the percentage. ‖ ∻Answer∻ ⚹ [Insert final answer here] ⚹ ∻Key Concept∻ ⚹ Ideal gas law and stoichiometry ⚹ ∻Explanation∻ ⚹ The volume of hydrogen gas produced in the reaction is used to determine the moles of zinc that reacted, which in turn is used to calculate the percentage of zinc in the alloy. ⚹◊What is the balanced chemical equation for the reaction between the Cu-Zn alloy and HCl acid to produce hydrogen gas?⍭ Generate me a similar question◊

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