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Chapter 6: Problem 60

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 mL of a 0.100-M solution of iron(III) sulfate?

Short Answer

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

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between barium chloride (BaCl2) and iron(III) sulfate (Fe2(SO4)3) is: \[ BaCl_2(aq) + Fe_2(SO_4)_3(aq) \rightarrow 2BaSO_4(s) + 6FeCl_3(aq) \]
02

Converting volumes to moles

Given 100.0 mL of 0.100 M barium chloride and 100.0 mL of 0.100 M iron(III) sulfate, we can convert the volume to moles using the formula: Moles = Molarity × Volume (in liters) For barium chloride: Moles of BaCl2 = (0.100 mol/L) × (100.0 mL / 1000 mL/L) = 0.0100 mol For iron(III) sulfate: Moles of Fe2(SO4)3 = (0.100 mol/L) × (100.0 mL / 1000 mL/L) = 0.0100 mol
03

Finding the limiting reactant

In the balanced equation, we can see that 1 mole of BaCl2 reacts with 0.5 moles of Fe2(SO4)3: 1 mol BaCl2 : 0.5 mol Fe2(SO4)3 Divide the number of moles of each reactant by their respective stoichiometric coefficients: For BaCl2: 0.0100 mol / 1 = 0.0100 For Fe2(SO4)3: 0.0100 mol / 0.5 = 0.0200 Since 0.0100 < 0.0200, BaCl2 is the limiting reactant as it has the lowest ratio.
04

Calculate the moles of product (barium sulfate) formed

According to the balanced equation, 1 mole of BaCl2 produces 2 moles of barium sulfate: 1 mol BaCl2 --> 2 mol BaSO4 Using the limiting reactant (BaCl2), we can find the moles of barium sulfate produced: Moles of BaSO4 = (moles of BaCl2) × (2 moles BaSO4 / 1 mol BaCl2) = (0.0100 mol) × 2 = 0.0200 mol
05

Calculate the mass of barium sulfate formed

Using the molar mass of barium sulfate (BaSO4), we can convert moles to mass: Molar mass of BaSO4 = 233.43 g/mol Mass of BaSO4 = Moles of BaSO4 × Molar mass of BaSO4 = (0.0200 mol) × (233.43 g/mol) = 4.6686 g Therefore, the mass of barium sulfate that can be produced when 100.0 mL of 0.100 M barium chloride is mixed with 100.0 mL of 0.100 M iron(III) sulfate is approximately 4.67 g.

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