QuestionJuly 30, 2025

The reaction of 0 .194 g of Zn with HCl(aq) gave 80.3 mL (at 23.4^circ C of hydrogen gas collected over water in a gas tube. The water was 18.5 mm higher in the gas tube than it was in the beaker. The barometer reading is 698. mm Hg. Following the steps on pages 1 to 3, complete this table of data. Volume of H_(2) Gas Generated (mL) square

The reaction of 0 .194 g of Zn with HCl(aq) gave 80.3 mL (at 23.4^circ C of hydrogen gas collected over water in a gas tube. The water was 18.5 mm higher in the gas tube than it was in the beaker. The barometer reading is 698. mm Hg. Following the steps on pages 1 to 3, complete this table of data. Volume of H_(2) Gas Generated (mL) square
The reaction of 0 .194 g of Zn with HCl(aq) gave 80.3 mL (at 23.4^circ C of
hydrogen gas collected over water in a gas tube. The water was 18.5 mm
higher in the gas tube than it was in the beaker. The barometer reading is
698. mm Hg. Following the steps on pages 1 to 3, complete this table of
data.
Volume of H_(2) Gas Generated (mL)
square

Solution
3.0(229 votes)

Answer

71.7 mL Explanation 1. Calculate the pressure of dry hydrogen gas Use the barometer reading and adjust for water vapor pressure. The pressure of dry hydrogen is given by P_{\text{dry}} = P_{\text{barometer}} - P_{\text{water}}. Convert the height difference of water to mm Hg using the density of mercury (13.6 g/cm³). P_{\text{water}} = \frac{18.5}{13.6} \approx 1.36 mm Hg. Therefore, P_{\text{dry}} = 698 - 1.36 = 696.64 mm Hg. 2. Correct volume for temperature and pressure Use the ideal gas law correction: V_1/T_1 = V_2/T_2, where T is in Kelvin. Convert 23.4^{\circ}C to Kelvin: T = 23.4 + 273.15 = 296.55 K. Assume standard conditions are T_0 = 273.15 K and P_0 = 760 mm Hg. Thus, V_2 = V_1 \times \frac{P_{\text{dry}}}{P_0} \times \frac{T_0}{T}. 3. Calculate corrected volume Given V_1 = 80.3 mL, calculate V_2: V_2 = 80.3 \times \frac{696.64}{760} \times \frac{273.15}{296.55} \approx 71.7 mL.

Explanation

1. Calculate the pressure of dry hydrogen gas<br /> Use the barometer reading and adjust for water vapor pressure. The pressure of dry hydrogen is given by $P_{\text{dry}} = P_{\text{barometer}} - P_{\text{water}}$. Convert the height difference of water to mm Hg using the density of mercury (13.6 g/cm³). $P_{\text{water}} = \frac{18.5}{13.6} \approx 1.36$ mm Hg. Therefore, $P_{\text{dry}} = 698 - 1.36 = 696.64$ mm Hg.<br /><br />2. Correct volume for temperature and pressure<br /> Use the ideal gas law correction: $V_1/T_1 = V_2/T_2$, where $T$ is in Kelvin. Convert $23.4^{\circ}C$ to Kelvin: $T = 23.4 + 273.15 = 296.55$ K. Assume standard conditions are $T_0 = 273.15$ K and $P_0 = 760$ mm Hg. Thus, $V_2 = V_1 \times \frac{P_{\text{dry}}}{P_0} \times \frac{T_0}{T}$.<br /><br />3. Calculate corrected volume<br /> Given $V_1 = 80.3$ mL, calculate $V_2$: $V_2 = 80.3 \times \frac{696.64}{760} \times \frac{273.15}{296.55} \approx 71.7$ mL.
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