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Chapter 4: Problem 64

What mass of barium sulfate can be produced 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?

Short Answer

Expert verified
The mass of barium sulfate produced 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 is 2.334 g.

Step by step solution

01

Convert volumes to moles of reactants

We are given the volumes (100.0 mL) and concentrations (0.100 M) of both reactants, so we can calculate the moles of each reactant by using the formula: moles = Molarity × Volume (in Liters) Moles of barium chloride (BaCl_2): moles = 0.100 mol/L × 100.0 mL × (1 L / 1000 mL) = 0.0100 mol Moles of iron(III) sulfate (Fe_2(SO_4)_3): moles = 0.100 mol/L × 100.0 mL × (1 L / 1000 mL) = 0.0100 mol
02

Identify the limiting reactant

In order to determine the limiting reactant, we must check how many moles of the other reactant are needed for each reactant according to the balanced chemical equation. From the equation: 1 mol BaCl_2 reacts with 1/2 mol Fe_2(SO_4)_3 Moles of Fe_2(SO_4)_3 needed for BaCl_2 = (1/2) × 0.0100 mol = 0.0050 mol Since we have 0.0100 mol of Fe_2(SO_4)_3, there is enough iron(III) sulfate to fully react with barium chloride. Therefore, barium chloride (BaCl_2) is the limiting reactant.
03

Calculate moles of barium sulfate produced

Now that we know the limiting reactant, we can determine the moles of barium sulfate (BaSO_4) produced using the stoichiometry of the balanced equation. From the equation: 1 mol BaCl_2 reacts with 1 mol of BaSO_4 Moles of BaSO_4 produced = 1 × moles of BaCl_2 = 1 × 0.0100 mol = 0.0100 mol
04

Calculate mass of barium sulfate

Now that we have the moles of barium sulfate (BaSO_4) produced, we can calculate its mass by using the molar mass: mass = moles × molar mass Molar mass of BaSO_4 = 137.33 (Ba) + 32.07 (S) + 4 × 16.00 (O) = 233.40 g/mol Mass of BaSO_4 produced = 0.0100 mol × 233.40 g/mol = 2.334 g The mass of barium sulfate produced 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 is 2.334 g.

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Most popular questions from this chapter

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ b. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)$ c. $\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)$ d. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)$

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Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. $$\operatorname{copper}(\mathrm{II}) \text { sulfate }(a q)+\mathrm{iron}(s) \longrightarrow (s)+\operatorname{iron}(\mathrm{II}) \text { sulfate }(a q)$$ $$\operatorname{copper}(\mathrm{II}) \text { sulfate }(a q)+\mathrm{iron}(s) \longrightarrow (s)+\text { iron (III) sulfate }(a q) $$ You place 87.7 mL of a 0.500-M solution of copper(II) sulfate in a beaker. You then add 2.00 g of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate 2.27 g of copper. Which equation above describes the reaction that occurred? Support your answer

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