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. 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 \(30 .^{\circ} \mathrm{C}\).)

In order to find the partial pressure of hydrogen gas, we can use Dalton's law of partial pressures, which states that the total pressure of a gas mixture is equal to the sum of its partial pressures: $$P_{\text{total}} = P_{\text{H}_{2}} + P_{\text{H} _2 \text{O}}$$ We are given the total pressure (\(P_{\text{total}} = 1.032 \text{ atm}\)) and the vapor pressure of water at 30°C (\(P_{\text{H}_{2} \text{O}} = 32 \text{ torr}\)). We can convert \(P_{\text{H}_{2} \text{O}}\) to atm: $$P_{\text{H}_{2} \text{O}} = \frac{32 \text{ torr}}{760 \text{ torr/atm}} = 0.0421 \text{ atm}$$ Now we can find the partial pressure of hydrogen gas by using Dalton's law: $$P_{\text{H}_{2}} = P_{\text{total}} - P_{\text{H}_{2} \text{O}}$$

Now we can substitute the pressure values into the equation and solve for the partial pressure of hydrogen gas: $$P_{\text{H}_{2}} = 1.032 \text{ atm} - 0.0421 \text{ atm} = 0.9899 \text{ atm}$$ The partial pressure of hydrogen gas is 0.9899 atm.

Now we need to find the number of moles of hydrogen gas to determine the amount of zinc reacted. We can use the Ideal Gas Law: $$PV = nRT$$ Rearranging the equation to find the number of moles (n): $$ n = \frac{PV}{RT}$$ We are given the volume (V) of hydrogen gas (240 mL = 0.240 L) and the temperature (T) is given in Celsius, but we need to convert it to Kelvin: $$T_K = 30°C + 273.15 = 303.15 \text{K}$$ The gas constant (R) in atm L/mol K is 0.0821, and we calculated the partial pressure of hydrogen gas (P) in step 2 as 0.9899 atm. Now we can plug in the values: $$n = \frac{(0.9899 \text{ atm})(0.240 \text{ L})}{(0.0821 \text{ atm L/mol K})(303.15 \text{K})}$$

Solve for the number of moles by substituting the values into the equation: $$n = \frac{(0.9899)(0.240)}{(0.0821)(303.15)} = 0.00994 \text{ mol}$$ The number of moles of hydrogen gas is 0.00994 mol.

Using the balanced equation given in the exercise, we can determine the number of moles of zinc required to produce the amount of hydrogen gas: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ From the balanced equation, we can see that 1 mol of Zn reacts to produce 1 mol of H2. Therefore, the number of moles of zinc required is equal to the number of moles of hydrogen gas: $$\text{moles of Zn} = \text{moles of H}_{2} = 0.00994 \text{ mol}$$

Finally, we can find the mass of zinc required to react by multiplying the number of moles of zinc by the molar mass of zinc: $$\text{mass of Zn} = \text{moles of Zn} \times \text{molar mass of Zn}$$ The molar mass of zinc is 65.38 g/mol. Therefore: $$\text{mass of Zn} = 0.00994 \text{ mol} \times 65.38 \frac{\text{g}}{\text{mol}}$$

Multiply the number of moles by the molar mass to find the mass of zinc: $$\text{mass of Zn} = 0.00994 \times 65.38 = 0.649 \text{ g}$$ The mass of zinc required to produce this quantity of hydrogen gas is 0.649 g.