Question

A 300.0 -g sample of a solid is made up of a uniform mixture of \(\mathrm{Na} \mathrm{NO}_{3}, \mathrm{MgCl}_{2},\) and \(\mathrm{BaCl}_{2} .\) A 100.0 -g sample of the mixture is dissolved in water and treated with an excess of KOH. The precipitate from the reaction has a mass of \(13.47 \mathrm{~g} .\) The remaining \(200.0-\mathrm{g}\) sample is also dissolved in water and treated with an aqueous solution of \(\mathrm{AgNO}_{3} .\) The resulting precipitate has a mass of \(195.8 \mathrm{~g}\). What are the masses of \(\mathrm{NaNO}_{3}, \mathrm{MgCl}_{2},\) and \(\mathrm{BaCl}_{2}\) in the 300.0 -g sample?

Short answer

Answer: The masses of NaNO3, MgCl2, and BaCl2 in the 300-g sample are 121.20 g, 57.12 g, and 121.68 g, respectively.

Step-by-step solution

Mass of precipitates in the first and second samples

First, let's find the mass of the precipitates in the first 100-g sample and the second 200-g sample. For the 100-g sample treated with KOH, we have a precipitate of 13.47 g. For the 200-g sample treated with AgNO3, we have a precipitate of 195.8 g.

Identify the precipitates from the given reactions

Now let's identify the precipitates formed from the reactions. When the mixture is treated with KOH, the precipitate formed is Mg(OH)2, as Mg2+ ions from MgCl2 react with OH- ions from KOH. When the mixture is treated with AgNO3, the precipitates formed are AgCl, as Cl- ions from MgCl2 and BaCl2 react with Ag+ ions from AgNO3.

Use stoichiometry to find mass of compounds

We can now use stoichiometry principles to find the mass of the compounds in the mixture. For the 100-g sample: - Calculate moles of Mg(OH)2 formed in the precipitate: \(\frac{13.47 \mathrm{~g}}{58.32(1) + 16(2) + 1(2) \mathrm{~g/mol}} = 0.200 \mathrm{~mol}\) - This means 0.200 mol of MgCl2 was present in the first 100-g sample: \(0.200 \mathrm{~mol} \times \frac{95.21 \mathrm{~g/mol}}{1 \mathrm{~mol}} = 19.04 \mathrm{~g}\) - Since the first 100-g sample is one-third of the original 300-g sample, the total mass of MgCl2 in the 300-g sample is 3 times the mass found in the 100-g sample: \(19.04 \mathrm{~g} \times 3 = 57.12 \mathrm{~g}\) For the 200-g sample: - Calculate moles of AgCl formed in the precipitate: \(\frac{195.8 \mathrm{~g}}{107.87(1) + 35.45(1) \mathrm{~g/mol}} = 1.200 \mathrm{~mol}\) - This means 1.200 mol of Cl- ions are present in the second 200-g sample - Since we already know the mass of MgCl2 in the 300-g sample, we can find the moles of Cl- ions associated with MgCl2 in the 200-g sample: \(\frac{57.12 \mathrm{~g}}{95.21 \mathrm{~g/mol}} = 0.600 \mathrm{~mol}\) - The remaining 0.600 mol of Cl- ions must be associated with BaCl2 in the second 200-g sample: \(0.600 \mathrm{~mol} \times \frac{137.33(1) + 35.45(2)\mathrm{~g/mol}}{1\mathrm{~mol}} = 121.68 \mathrm{~g} \)

Calculate mass of NaNO3

Subtract the sum of MgCl2 and BaCl2 from the initial 300-g sample mass to find the mass of NaNO3: \(300.0 \mathrm{~g} - (57.12 \mathrm{~g} + 121.68 \mathrm{~g}) = 121.20 \mathrm{~g}\) So, the masses of NaNO3, MgCl2, and BaCl2 in the 300-g sample are: NaNO3: 121.20 g MgCl2: 57.12 g BaCl2: 121.68 g