A \(0.5895-\mathrm{g}\) sample of impure magnesium hydroxide is dissolved in \(100.0 \mathrm{~mL}\) of \(0.2050 \mathrm{MHCl}\) solution. The excess acid then needs \(19.85 \mathrm{~mL}\) of \(0.1020 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

To determine the moles of HCl consumed in the reaction with Mg(OH)₂, first find out how many moles of HCl reacted with the excess NaOH. The balanced reaction between NaOH and HCl is: NaOH + HCl → NaCl + H₂O The stoichiometry of this reaction is 1:1, which means that one mole of NaOH reacts with one mole of HCl. Use the formula: moles = Molarity * Volume (in L) Find the moles of NaOH used to neutralize excess HCl: moles(NaOH) = 0.1020 * (19.85/1000) L = 0.0020257 mol Since the stoichiometry is 1:1, the same amount of HCl reacted with this NaOH: moles(HCl reacted with NaOH) = moles(NaOH) = 0.0020257 mol Now find the moles of HCl initially present in the solution: moles(HCl initially) = 0.2050 * (100/1000) L = 0.0205 mol Now find the moles of HCl that reacted with magnesium hydroxide in the sample: moles(HCl reacted with Mg(OH)₂) = moles(HCl initially) - moles(HCl reacted with NaOH) moles(HCl reacted with Mg(OH)₂) = 0.0205 - 0.0020257 = 0.0184743 mol

Next, we need to find out how many moles of magnesium hydroxide reacted with the HCl. The balanced equation for the reaction between Mg(OH)₂ and HCl is: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O The stoichiometry of this reaction is 1:2, meaning one mole of Mg(OH)₂ reacts with two moles of HCl. Use the stoichiometry to calculate the moles of Mg(OH)₂ in the sample: moles(Mg(OH)₂) = moles(HCl reacted with Mg(OH)₂) / 2 moles(Mg(OH)₂) = 0.0184743 / 2 = 0.00923715 mol

To find the mass of magnesium hydroxide, multiply the moles of Mg(OH)₂ with its molar mass. First, find the molar mass of Mg(OH)₂: Molar mass(Mg(OH)₂) = 24.31 (Mg) + 2(16.00 + 1.01) (2OH) = 58.32 g/mol Now, calculate the mass of Mg(OH)₂ in the sample: mass(Mg(OH)₂) = moles(Mg(OH)₂) * molar mass(Mg(OH)₂) mass(Mg(OH)₂) = 0.00923715 * 58.32 g/mol = 0.538288 g

Finally, calculate the percentage by mass of magnesium hydroxide in the impure sample using the formula: percentage by mass = (mass of Mg(OH)₂ / mass of impure sample) * 100 percentage by mass = (0.538288 / 0.5895) * 100 = 91.31% The percentage by mass of magnesium hydroxide in the impure sample is approximately 91.31%.