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Chapter 4: Problem 58

What mass of barium sulfate can be produced when \(100.0 \mathrm{~mL}\) of a \(0.100 M\) solution of barium chloride is mixed with \(100.0\) \(\mathrm{mL}\) of a \(0.100 \mathrm{M}\) solution of iron(III) sulfate?

Short Answer

Expert verified
When 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron(III) sulfate, 4.67 g of barium sulfate can be produced.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between barium chloride (BaCl2) and iron(III) sulfate (Fe2(SO4)3) is: \[ 3BaCl_2(aq) + 2Fe_2(SO_4)_3(aq) \rightarrow 6BaSO_4(s) + 4FeCl_3(aq) \] In this reaction, three moles of barium chloride react with two moles of iron(III) sulfate to form six moles of barium sulfate and four moles of iron(III) chloride.
02

Calculate the moles of each reactant

We have 100.0 mL of a 0.100 M solution of barium chloride, and 100.0 mL of a 0.100 M solution of iron(III) sulfate. To calculate the moles of each reactant, use the formula: Moles = Molarity × Volume (in L) For barium chloride: Moles = 0.100 mol/L × 0.100 L = 0.0100 mol For iron(III) sulfate: Moles = 0.100 mol/L × 0.100 L = 0.0100 mol
03

Determine the limiting reactant

To determine the limiting reactant, we'll compare the moles of each reactant to their stoichiometric ratio in the balanced chemical equation. For barium chloride (3 moles of BaCl2 needed for each 2 moles of Fe2(SO4)3): \( \frac{0.0100~mol}{3} = 0.00333 \) For iron(III) sulfate: \( \frac{0.0100~mol}{2} = 0.00500 \) Since the value for barium chloride is smaller, barium chloride is the limiting reactant.
04

Calculate the moles of the product

In the balanced chemical equation, three moles of barium chloride react to produce six moles of barium sulfate (BaSO4). So, we can use the stoichiometric ratio to find the moles of barium sulfate produced: Moles of barium sulfate = \( \frac{6}{3} \times 0.0100~mol \) = 0.0200 mol
05

Convert moles of the product to mass

Finally, to convert the moles of barium sulfate to mass, use the formula: Mass = Moles × Molar mass The molar mass of barium sulfate (BaSO4) is: \[ 137.33 (Ba) + 32.07 (S) + 15.999 (O) \times 4 = 233.43~g/mol\] So, the mass of barium sulfate produced is: Mass = 0.0200 mol × 233.43 g/mol = 4.67 g Therefore, when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron(III) sulfate, 4.67 g of barium sulfate can be produced.

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Most popular questions from this chapter

A solution was prepared by mixing \(50.00 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) and \(100.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the molarity of the final solution of nitric acid.

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0\) \(\mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

Write net ionic equations for the reaction, if any, that occurs when aqueous solutions of the following are mixed. a. ammonium sulfate and barium nitrate b. lead(II) nitrate and sodium chloride c. sodium phosphate and potassium nitrate d. sodium bromide and rubidium chloride e. copper(II) chloride and sodium hydroxide

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