Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose \(240 . \mathrm{mL}\) of hydrogen gas is collected at \(30 .{ }^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

Step by step solution

01

Calculate the partial pressure of water vapor

First, we need to convert the vapor pressure of water, given in torr, to atm. We know that 1 atm = 760 torr. So: P_water (atm) = \( \frac{32 \, \text{torr}}{760 \, \text{torr/atm}} \)= 0.042 atm

02

Calculate the partial pressure of hydrogen gas

Now, we will use Dalton's Law of Partial Pressures to find the partial pressure of hydrogen gas: P_total = P_H2 + P_water P_H2 = P_total - P_water P_H2 = 1.032 atm - 0.042 atm = 0.990 atm

03

Find the moles of hydrogen gas

Next, we will use the Ideal Gas Law to find the moles of hydrogen gas. The Ideal Gas Law is given by: PV = nRT In this case, we know the volume (V) is 240 mL, which needs to be converted to L: V = 240 mL × \( \frac{1\, \text{L}}{1000\, \text{mL}} \) = 0.240 L We also know the temperature (30°C) must be converted to Kelvin (K) to use in the Ideal Gas Law: T = 30°C + 273.15 = 303.15 K We can now find the moles of hydrogen gas (n) using the ideal gas law formula: n = \( \frac{PV}{RT} \) n_H2 = \( \frac{(0.990\, \text{atm})(0.240\, \text{L})}{(0.08206\, \text{L atm/mol K})(303.15\, \text{K})} \) = 0.00986 mol

04

Calculate the mass of zinc

Finally, we will use stoichiometry to find the amount of zinc required for the reaction. From the balanced chemical equation, we have: Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) 1 mole of Zn reacts to produce 1 mole of H2. Therefore, the moles of Zn required are equal to the moles of H2 produced: n_Zn = n_H2 = 0.00986 mol Now, we can calculate the mass of zinc required using its molar mass (65.38 g/mol): mass_Zn = n_Zn × molar mass of Zn mass_Zn = 0.00986 mol × 65.38 g/mol = 0.644 g So, the partial pressure of hydrogen gas is 0.990 atm, and the amount of zinc required to produce this quantity of hydrogen gas is 0.644 g.

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