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Chapter 4: Problem 134

A 10.00-g sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead(II) nitrate to form 21.75 g of solid. Determine the mass percent of sodium chloride in the original mixture.

Short Answer

Expert verified
The mass percent of sodium chloride in the original mixture is 19.37%.

Step by step solution

01

Write balanced chemical equations for the reactions

First, write balanced chemical equations for the reactions of sodium chloride and potassium sulfate with lead(II) nitrate. The reactants are: Sodium chloride (NaCl) Potassium sulfate (K2SO4) Lead(II) nitrate (Pb(NO3)2) The reactions are: NaCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + NaNO3(aq) K2SO4(aq) + 2Pb(NO3)2(aq) → 2PbSO4(s) + 2KNO3(aq)
02

Calculate the moles of the solid formed

Next, calculate the moles of the solid formed after the reaction. We know the mass of the solid formed and we can find the molar mass of the possible solids to determine the moles of each possible solid formed. For PbCl2: molar mass \(= 207.2 + 2 \times 35.5 = 278.2 g/mol\) For PbSO4: molar mass \(= 207.2 + 32 + 4 \times 16 = 303.2 g/mol\) Let x be the mass (in grams) of PbCl2 formed and y be the mass (in grams) of PbSO4 formed. We know the total mass of the solid formed is 21.75 g, so: x + y = 21.75
03

Calculate the moles of the initial components in the mixture

Now, convert the mass of each solid formed to moles using their molar mass: Moles of PbCl2: \(\frac{x}{278.2}\) Moles of PbSO4: \(\frac{y}{303.2}\) Since sodium chloride and potassium sulfate react in a 1:1 and 1:2 ratio with lead(II) nitrate respectively, the moles of sodium chloride and potassium sulfate in the original mixture are equal to the moles of PbCl2 and half the moles of PbSO4, respectively. Moles of NaCl: \(\frac{x}{278.2}\) Moles of K2SO4: \(\frac{y}{2 \times 303.2}\)
04

Calculate the mass of the original components in the mixture

Next, calculate the mass of sodium chloride (NaCl) and potassium sulfate (K2SO4) in the original mixture using their molar mass. For NaCl: molar mass \(= 22.99 + 35.45 = 58.44 g/mol\) For K2SO4: molar mass \(= 2 \times 39.10 + 32.07 + 4 \times 16.00 = 174.26 g/mol\) Mass of NaCl: \(58.44 \times \frac{x}{278.2}\) Mass of K2SO4: \(174.26 \times \frac{y}{2 \times 303.2}\)
05

Determine the mass percent of sodium chloride

We're given that the total mass of the original mixture is 10.00 g. Therefore: \(58.44 \times \frac{x}{278.2} + 174.26 \times \frac{y}{2 \times 303.2} = 10\) Solving for x and y, we have: x = 9.237 g (mass of PbCl2) y = 12.51 g (mass of PbSO4) Now, find the mass of NaCl in the original mixture: Mass of NaCl: \(58.44 \times \frac{9.237}{278.2} = 1.937 g\) Finally, calculate the mass percent of sodium chloride in the original mixture: Mass percent of NaCl: \(\frac{1.937}{10} \times 100 = 19.37\%\) So, the mass percent of sodium chloride in the original mixture is 19.37%.

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