(a) A sample of hydrogen gas is generated in a closed container by reacting \(1.750 \mathrm{~g}\) of zinc metal with \(50.0 \mathrm{~mL}\) of $1.00 \mathrm{M}$ hydrochloric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 150 mL. Calculate the partial pressure of the hydrogen gas in this volume at $25^{\circ} \mathrm{C}$, ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is $7.7 \times 10^{-6} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}$. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?

Step by step solution

01

Write the balanced chemical equation

The reaction between zinc metal and hydrochloric acid can be written as: \[Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)\]

02

Calculate the number of moles of zinc and HCl

First, we need to find the number of moles of zinc and hydrochloric acid involved in the reaction. To do this, we will use the given mass of zinc and the concentration and volume of the hydrochloric acid. Moles of Zn: \(n_{Zn} = \frac{mass_{Zn}}{molar~mass_{Zn}} = \frac{1.750~g}{65.38~g/mol} = 0.0268~mol\) Moles of HCl: \(n_{HCl} = C_{HCl} \times V_{HCl} = 1.00~M \times 0.0500~L = 0.0500~mol\)

03

Determine the limiting reactant and calculate moles of hydrogen gas

To find the limiting reactant, we will compare the mole ratios of the reactants to the balanced chemical equation. In the balanced equation, we see that 1 mole of Zn reacts with 2 moles of HCl. Therefore, we will divide the moles of each reactant by their corresponding stoichiometric coefficients and compare: \(\frac{n_{Zn}}{1} = 0.0268~mol\) \(\frac{n_{HCl}}{2} = 0.0250~mol\) Since the ratio for HCl is smaller, HCl is the limiting reactant. Now we can calculate the number of moles of hydrogen gas produced using the stoichiometry of the balanced equation: \(n_{H_2} = \frac{1}{2} \times n_{HCl} = \frac{1}{2} \times 0.0500~mol = 0.0250~mol\)

04

Calculate the partial pressure of hydrogen gas

To find the partial pressure of hydrogen gas, we will use the ideal gas law, \[PV = nRT\], 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. In this case, we have: \(V = 150~mL = 0.150~L\) \(T = 25^{\circ}C + 273.15 = 298.15~K\) \(R = 0.08206~L\cdot atm/mol\cdot K\) Now let's solve for the partial pressure of hydrogen gas: \(P_H = \frac{n_H RT}{V} = \frac{0.0250~mol \times 0.08206~L\cdot atm/mol\cdot K \times 298.15~K}{0.150~L} = 4.10~atm\)

05

Calculate the moles of dissolved hydrogen gas

By using Henry's law constant for the hydrogen gas, we can estimate the number of moles of hydrogen gas dissolved in the solution. Henry's law can be described by the equation: \[C = K_HP\] Where \(C\) is the dissolved gas concentration in mol/m³, \(K_H\) is Henry's law constant, and \(P\) is the partial pressure of the gas. In this case, we have: \(K_H = 7.7 \times 10^{-6}~mol/(m³Pa)\) \(P = 4.10~atm \rightarrow 4.10~atm \times 101325~Pa/atm = 415,233~Pa\) Now let's solve for the concentration of dissolved hydrogen gas: \(C = K_HP = 7.7 \times 10^{-6} \times 415,233 = 3.2~mol/m³\) Since the volume of the solution is \(0.150~L\) or \(1.5 \times 10^{-4}~m³\), the number of moles of dissolved hydrogen gas can be calculated as: \(n_{dissolved} = C \times V = 3.2~mol/m³ \times 1.5 \times 10^{-4}~m³ = 4.8 \times 10^{-4}~mol\)

06

Calculate the fraction of dissolved hydrogen gas and evaluate the assumption

Next, calculate the fraction of dissolved hydrogen gas: \(\text{Fraction of dissolved hydrogen gas} = \frac{n_{dissolved}}{n_{H_2}} = \frac{4.8 \times 10^{-4}~mol}{0.0250~mol} = 1.9 \times 10^{-2}\) This fraction is equal to about 1.9%, which indicates that only a small portion of the hydrogen gas is dissolved in the solution. Therefore, it was reasonable to ignore the dissolved hydrogen gas in part (b) for this particular problem.

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