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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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Although many sulfate salts are soluble in water, calcium sulfate is not (Table 7. 1). Therefore, a solution of calcium chloride will react with sodium sulfate solution to produce a precipitate of calcium sulfate. The balanced equation is

:math> CaCl 2 ( a q ) + Na 2 SO 4 ( a q ) CaSO 4 ( s ) + 2 NaCl ( a q )

a solution containing 5.21 g of calcium chloride is combined with a solution containing 4.95 g of sodium sulfate, which is the limiting reactant? Which reactant is present in excess?

Interpretation Introduction

Interpretation:

The limiting reactant and reactant present in excess should be calculated.

Concept Introduction:

The expected or theoretical yield of the product depends on the limiting reactant of the reaction. Limiting reactant is the reactant that limits the mass of product formed in the reaction. The reactant other than limiting reactant is known as excess reactant.

Explanation

The balanced chemical reaction of solution of calcium chloride with sodium sulphate solution produces a precipitate of calcium sulphate is as follows:

CaCl2(aq)+Na2SO4(aq)CaSO4(s)+2NaCl(aq)

The mass of calcium chloride is 5.21 g and that of sodium sulphate is 4.95 g.

To determine the limiting reactant, number of moles of each reactant should be calculated.

Mass of calcium chloride is 5.21 g and molar mass is 110.98 g/mol thus, number of moles will be:

n=mM

Putting the values,

n=5.21 g110.98 g/mol=0.047 mol

From the balanced chemical reaction, 1 mol of calcium chloride produces 1 mol of calcium sulphate thus, 0.047 mol produces 0.047 mol of calcium sulphate.

Mass of calcium sulphate is calculated as follows:

m=n×M

Molar mass of calcium sulphate is 136

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