Question

When water is added to a mixture of aluminum metal and sodium hydroxide, hydrogen gas is produced. This is the reaction used in commercial drain cleaners: \(2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)+3 \mathrm{H}_{2}(g)\) A sufficient amount of water is added to \(49.92 \mathrm{~g}\) of \(\mathrm{NaOH}\) to make \(0.600 \mathrm{~L}\) of solution; \(41.28 \mathrm{~g}\) of \(\mathrm{Al}\) is added to this solution and hydrogen gas is formed. (a) Calculate the molarity of the initial \(\mathrm{NaOH}\) solution. (b) How many moles of hydrogen were formed? (c) The hydrogen was collected over water at \(25^{\circ} \mathrm{C}\) and \(758.6 \mathrm{~mm} \mathrm{Hg}\). The vapor pressure of water at this temperature is \(23.8 \mathrm{~mm} \mathrm{Hg}\). What volume of hydrogen was generated?

Short answer

a) The molarity of the initial NaOH solution is 2.080 M. b) The number of moles of hydrogen formed is 2.294 moles. c) The volume of hydrogen gas collected, considering the effect of water vapor pressure, is 57.3 L.

Step-by-step solution

Calculate the molarity of the initial NaOH solution

To find the molarity of the NaOH solution, we will first find the moles of NaOH and then divide by the liters of the solution. Moles of NaOH = mass of NaOH / molar mass of NaOH Molarity (NaOH) = moles of NaOH / volume of solution (in Liters) The molar mass of NaOH is 22.99 g/mol (for Na) + 15.999 g/mol (for O) + 1.007 g/mol (for H) = 39.996 g/mol. Moles of NaOH = 49.92 g / 39.996 g/mol = 1.248 mol Molarity (NaOH) = 1.248 mol / 0.600 L = 2.080 M

Calculate the number of moles of hydrogen formed

Using the balanced chemical equation, we can find the moles of hydrogen formed by applying stoichiometry: $2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)+3 \mathrm{H}_{2}(g)$ From the equation, 2 moles Al react with 2 moles OH\(^-\) to give 3 moles of H\(_2\). Find moles of Al: 41.28 g Al / 26.98 g/mol (molar mass of Al) = 1.529 mol Al Since the aluminum is the limiting reactant, we can determine the moles of hydrogen produced: Moles of H\(_2\) = (3 moles H\(_2\) / 2 moles Al) × 1.529 moles Al = 2.294 moles H\(_2\)

Calculate the volume of hydrogen gas collected

We are now given the conditions under which the hydrogen gas was collected: \(T = 25^{\circ} \text{C}\), \(P_\text{total} = 758.6 \text{ mm Hg}\), and the vapor pressure of water, \(P_\text{water} = 23.8 \text{ mm Hg}\). We can use the Ideal Gas Law to find the volume of hydrogen gas collected. First, let's find the pressure of hydrogen gas by subtracting the vapor pressure of water from the total pressure: \(P_\text{H_2} = P_\text{total} - P_\text{water} = 758.6 \text{ mm Hg} - 23.8 \text{ mm Hg} = 734.8 \text{ mm Hg}\) Now, convert temperature to Kelvin: \(T = 25^{\circ} \text{C} + 273.15 = 298.15 \text{ K}\) Convert pressure from mm Hg to atm: \(P_\text{H_2} = 734.8 \text{ mm Hg} \times \frac{1 \text{ atm}}{760 \text{ mm Hg}} = 0.967 \text{ atm}\) Use the Ideal Gas Law: PV = nRT, solve for V: \(V = \frac{nRT}{P} = \frac{2.294 \text{ mol} \times 0.0821 \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}} \times 298.15 \text{K}}{0.967 \text{ atm}} = 57.3 \text{ L}\) The volume of hydrogen gas collected is 57.3 L.