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?

First, we need to calculate the moles of NaOH and Al using their given masses. To do this, we will use the molar masses of NaOH and Al: NaOH: 1 Na (22.99 g/mol) + 1 O (16.00 g/mol) + 1 H (1.01 g/mol) = 40.00 g/mol Al: 1 Al (26.98 g/mol) Now, we can find the moles of each substance by dividing their given mass by their molar mass. Moles of NaOH = \(\frac{49.92 \mathrm{~g}}{40.00 \mathrm{~g/mol}} = 1.248 \mathrm{~mol}\) Moles of Al = \(\frac{41.28 \mathrm{~g}}{26.98 \mathrm{~g/mol}} = 1.53 \mathrm{~mol}\)

Now, we can calculate the molarity of NaOH solution by dividing the moles of NaOH by the volume of the solution in liters. Moles of NaOH = 1.248 mol Volume of solution = 0.600 L Molarity of NaOH = \(\frac{1.248 \mathrm{~mol}}{0.600 \mathrm{~L}} = 2.080 \mathrm{~M}\) The initial molarity of the NaOH solution is 2.080 M.

To find the number of moles of hydrogen formed, we will use the balanced chemical equation: $$ 2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{Al}(\mathrm{OH})^{4-}(a q)+3 \mathrm{H}_{2}(g) $$ From the equation, we can see that 2 moles of Al react with 2 moles of hydroxide ions (OH⁻) present in NaOH to produce 3 moles of hydrogen gas. Therefore, we will first find out the limiting reactant, which is the reactant that runs out first and will limit the amount of hydrogen gas formed. The given moles of Al are 1.53 mol and moles of NaOH (providing OH⁻) are 1.248 mol. As the stoichiometry ratio is 2:2 (1:1), we can see that there will be an excess of Al, but NaOH will be the limiting reactant. Therefore, we will use the number of moles of NaOH to find the number of moles of hydrogen produced. Since the stoichiometry ratio between NaOH and hydrogen is 2:3, the number of moles of hydrogen formed will be: Moles hydrogen = \(\frac{3}{2}\) * Moles NaOH = \(\frac{3}{2} * 1.248 \mathrm{~mol} = 1.872 \mathrm{~mol}\) There are 1.872 moles of hydrogen formed.

We are given that hydrogen gas is collected over water at 25°C (298 K) and 758.6 mmHg. The vapor pressure of water at 25°C is 23.8 mmHg. We will first convert the pressure into atm by dividing it by 760, and then subtract the vapor pressure of water to get the pressure of hydrogen gas. Total pressure = \(\frac{758.6 \mathrm{~mmHg}}{760 \mathrm{~mmHg/atm}} = 0.998 \mathrm{~atm}\) Pressure of hydrogen = Total pressure - Vapor pressure of water = \(0.998 \mathrm{~atm} - \frac{23.8 \mathrm{~mmHg}}{760 \mathrm{~mmHg/atm}} = 0.998 \mathrm{~atm} - 0.031 \mathrm{~atm} = 0.967 \mathrm{~atm}\) Now, we can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is number of moles, R is the gas constant (0.0821 L atm/mol K), and T is temperature. We are solving for volume (V). \(V = \frac{nRT}{P} = \frac{1.872 \mathrm{~mol} * 0.0821 \mathrm{~L~atm/mol~K} * 298 \mathrm{~K}}{0.967 \mathrm{~atm}} = 45.78 \mathrm{~L}\) The volume of hydrogen gas generated is 45.78 L.