Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 10, Problem 45E

A solution containing 1.63 g of barium chloride is added to a solution containing 2.40 g of sodium chromate (chromate ion, CrO 4 2 ). Find the mass in grams of barium chromate that can precipitate. Also determine which reactant was in excess, as well as the mass in grams in excess of the amount required by the limiting reactant.

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Interpretation Introduction

Interpretation:

The mass of precipitated barium chromate when barium chloride reacts with sodium chromate is to be calculated. The reactant and its mass which is present in excess in the reaction mixture is to be determined.

Concept introduction:

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limit the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer to Problem 45E

The mass of precipitated barium chromate when barium chloride reacts with sodium chromate is 1.98g. The reactant that is present in excess is sodium chromate. The mass of sodium chromate present in excess is 1.13g.

Explanation of Solution

The reaction between barium chloride and sodium chromate is shown below.

BaCl2+Na2CrO42NaCl+BaCrO4

The mass of BaCl2 reacted is 1.63g.

The mass of Na2CrO4 reacted is 2.40g.

The molar mass of BaCl2 is 208.2g/mol.

The molar mass of Na2CrO4 is 161.98g/mol.

The molar mass of BaCrO4 is 253.3g/mol.

The number of moles of a substance is given by the expression shown below.

n=mM …(1)

Where,

m is the mass of the substance.

M is the molar mass of the substance.

Substitute the mass and molar mass of BaCl2 in the equation (1).

n=1.63g208.2g/mol=7.829×103mol

Therefore, the number of moles of BaCl2 present in the reaction mixture is 7.829×103mol.

Substitute the mass and molar mass of Na2CrO4 in the equation (1).

n=2.40g161.98g/mol=1.48×102mol

Therefore, the number of moles of Na2CrO4 present in the reaction mixture is 1.48×102mol.

One mole of BaCl2 reacts with one mole of Na2CrO4. Therefore, the relation between the number of moles of BaCl2 and Na2CrO4 is given by the expression shown below.

nBaCl2=nNa2CrO4 …(2)

Where,

nNa2CrO4 is the number of moles of Na2CrO4.

nBaCl2 is the number of moles of BaCl2.

Substitute the value of nNa2CrO4 in the equation (2).

nBaCl2=1.48×102mol

The required amount of BaCl2 is greater than the available amount of BaCl2. Therefore. BaCl2 is the limiting reagent. The sodium chromate is present in the excess.

One mole of BaCl2 produces one mole of BaCrO4. Therefore, the relation between the number of moles of BaCrO4 and BaCl2 is given by the expression shown below.

nBaCrO4=nBaCl2 …(3)

Where,

nBaCl2 is the number of moles of BaCl2.

nBaCrO4 is the number of moles of BaCrO4.

Substitute the value of nBaCl2 in the equation (3).

nCH3CH3=7.829×103mol

Rearrange the equation (1) for the value of m.

m=nM …(4)

Substitute the value of molar mass and the number of moles of BaCrO4 in the equation (4).

m=(7.829×103mol)(253.3g/mol)=1.98g

Therefore, the mass of precipitated barium chromate when barium chloride reacts with sodium chromate is 1.98g.

The number of moles of sodium chromate that will react with 7.829×103mol of barium chloride is 7.829×103mol.

Substitute the value of molar mass and the number of moles of sodium chromate in the equation (4).

m=(7.829×103mol)(161.98g/mol)=1.268g1.27g

Therefore, the mass of sodium chromate reacted to produce 1.98g of barium is 1.27g.

The excess mass of sodium chromate is given by the expression as shown below.

mNa2CrO4=m1m2

Where,

m1 is the initial mass of sodium chromate present in the reaction mixture.

m2 is the mass of sodium chromate reacted in the reaction.

Substitute the value of m1 and m2 in the above equation.

mNa2CrO4=2.40g1.27g=1.13g

Therefore, the mass of sodium chromate present in excess is 1.13g.

Conclusion

The mass of precipitated barium chromate when barium chloride reacts with sodium chromate is 1.98g. The reactant that is present in excess is sodium chromate. The mass of sodium chromate present in excess is 1.13g.

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Chapter 10 Solutions

Introductory Chemistry: An Active Learning Approach

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