What volume of H₂ would be collected at 22.0^oC and a pressure of 713 torr if 2.65 g of zinc react according to the equation: Zn(s)+ 2HCl(aq)→ ZnCl₂(aq)+ H₂(g) A)0.908 L B)0.0405 L C)1.04 L D)2.08 L E)0.523 L

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Quizplus · Accepted Answer

First, convert the given mass of zinc to moles using its molar mass (65.38 g/mol for Zn). $2.65 \, \text{g Zn} \times \frac{1 \, \text{mol Zn}}{65.38 \, \text{g Zn}} = 0.0405 \, \text{mol Zn}$. According to the balanced chemical equation, the mole ratio of Zn to $H_2$ is 1:1, so 0.0405 moles of Zn will produce 0.0405 moles of $H_2$. Use the ideal gas law $PV = nRT$ to find the volume of $H_2$ at the given conditions, converting pressure to atm (713 torr / 760 torr/atm = 0.938 atm) and temperature to Kelvin (22°C + 273 = 295 K). $V = \frac{nRT}{P} = \frac{0.0405 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 295 \, \text{K}}{0.938 \, \text{atm}} = 1.04 \, \text{L}$.

What Volume of H2 Would Be Collected at 22

What Volume of H₂ Would Be Collected at 22