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Chapter 3: Q3.106P (page 89)

How many grams of solid barium sulfate form when 35.0 mL of 0.160 M barium chloride reacts with 58.0 mL of 0.065 M sodium sulfate? Aqueous sodium chloride forms also.

Short Answer

Expert verified

When a 35.0 mL of 0.160 M barium chloride reacts with a 58.0 mL of 0.065 M sodium sulfate, 0.88 g of BaSO4will form.

Step by step solution

01

Calculating the number of moles of BaSO4

Calculate the mole number produced using the given reactants:

Moles of BaSO4=35×10-3L BaCl2soln×0.16 mol BaCl21 L BaCl2 soln×1 mol BaSO41 mol BaCl2                                       =5.6×10-3mol BaSO4                                       =58×10-3L NaSO4soln×0.065 mol NaSO41 L NaSO4 soln×1 mol BaSO41 mol NaSO4                                       =3.77×10-3mol BaSO4.

02

Calculating the mass of BaSO4

Multiply the molar mass:

Mass of BaSO4=3.77×10-3mol BaSO4×233.38 g BaSO41 mol BaSO4                                    =0.88 g BaSO4.

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