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Chapter 12 Solutions

12.6

Solutions in Chemical Reactions

When a BaCl2 solution is added to a Na2SO4 solution, BaSO4, a white solid, forms.

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Molarity in Chemical Reactions

In a chemical reaction,• the volume and molarity of a solution are used

to determine the moles of a reactant or product

volume (L) x molarity ( mol ) = moles

1 L• if molarity (mol/L) and moles are given, the

volume (L) can be determined

mol x 1 L = volume (L)

mol

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Calculations Involving Solutions in Chemical Reactions

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Example of Using Molarity in a Chemical Equation

How many milliliters of a 3.00 M HCl solution are needed to react with 4.85 g of CaCO3?

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

STEP 1 State the given and needed quantities. Given 3.00 M HCl solution; 4.85 g of CaCO3

Need volume in milliliters

STEP 2 Write a plan to calculate needed quantity or concentration.

grams of CaCO3 moles of CaCO3 moles of HCl milliliters of HCl

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Example of Using Molarity in a Chemical Equation (continued)

STEP 3 Write equalities and conversion factors including mole-mole and concentration factors.1 mol of CaCO3 = 100.09 g of CaCO3

1 mol CaCO3 and 100.09 g CaCO3 100.09 g CaCO3 1 mol CaCO3

1 mol of CaCO3 = 2 mol of HCl 1 mol CaCO3 and 2 mol HCl 2 mol HCl 1 mol CaCO3

1000 mL of HCl solution = 3.00 mol of HCl 1000 mL HCl solution and 3.00 mol HCl

3.00 mol HCl 1000 mL HCl solution

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Example of Using Molarity in a Chemical Equation (continued)

STEP 4 Set up problem to calculate needed quantity or concentration.

4.85 g CaCO3 x 1 mol CaCO3 x 2 mol HCl 100.09 g CaCO3 1 mol CaCO3

x 1000 mL HCl = 32.3 mL of HCl solution 3.00 mol HCl

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Learning Check

How many milliliters of a 0.150 M Na2S solution are

needed to react with 18.5 mL of a 0.225 M NiCl2

solution?

NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)

A. 4.16 mL

B. 6.24 mL

C. 27.8 mL

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Solution

STEP 1 State the given and needed quantities. Given 0.0185 L of a 0.225 M NiCl2 solution;

0.150 M Na2S solution

Need milliliters of Na2S solution

STEP 2 Write a plan to calculate needed quantity or concentration.

liters of NiCl2 solution moles of NiCl2 solution

moles of Na2S solution milliliters of Na2S solution

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Solution (continued)

STEP 3 Write equalities and conversion factors including mole-mole and concentration factors.

0.225 mol of NiCl2 = 1 L of NiCl2 solution 0.225 mol NiCl2 and 1 L NiCl2

1 L NiCl2 0.225 NiCl21 mol of NiCl2 = 1 mol of Na2S

1 mol NiCl2 and 1 mol Na2S 1 mol Na2S 1 mol NiCl2

1000 mL of Na2S solution = 0.150 mol of Na2S 1000 mL HCl and 0.150 mol HCl

0.150 mol HCl 1000 mL HCl

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Solution (continued)

STEP 4 Set up problem to calculate needed quantity or concentration.

0.0185 L x 0.225 mol NiCl2 x 1 mol Na2S x 1000 mL

1 L 1 mol NiCl2 0.150 mol

= 27.8 mL of Na2S solution (C)

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Learning Check

How many liters of H2 gas at STP are produced

when 6.25 g of Zn react with 20.0 mL of a 1.50 M

HCl solution?

Zn(s) + 2HCl(aq) ZnCl2 (aq) + H2(g)

A. 4.28 L of H2

B. 0.336 L of H2

C. 0.168 L of H2

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Solution

STEP 1 State the given and needed quantities.

Given 6.25 g of zinc; 0.0200 L of a 1.50 M HCl solution

Need L of H2 gas at STP

STEP 2 Write a plan to calculate needed quantity or concentration. Limiting reactant: lowest number of moles of H2

1) grams of zinc moles of zinc moles of H2

2) liters of HCl solution moles of HCl solution moles of H2

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Solution (continued)

STEP 3 Write equalities and conversion factors including mole-mole and concentration factors.

1 mol of Zn = 65.41 g of Zn 1 mol Zn and 65.41 g Zn 65.41 g Zn 1 mol Zn

1 mol of Zn = 1 mol of H2

1 mol Zn and 1 mol H2 1 mol H2 1 mol Zn

2 mol of HCl = 1 mol of H2 2 mol HCl and 1 mol H2 1 mol H2 2 mol HCl

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Solution (continued)

STEP 3 (continued)

1 mol of H2 = 22.4 L of H2 22.4 L H2 and 1 mol H2 1 mol H2 22.L H2

1.50 mol of HCl = 1 L of HCl solution 1.50 mol HCl and 1 L HCl solution 1 L HCl solution 1.50 mol HCl

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Solution (continued)

STEP 4 Set up problem to calculate needed quantity or concentration.

6.25 g Zn x 1.00 mol Zn x 1 mol H2 = 0.0956 mol of H2

65.41 g Zn 1 mol Zn0.0200 L x 1.50 mol HCl x 1 mol H2 = 0.0150 mol of H2

1 L 2 mol HCl (smaller)

Using the smaller number of moles of H2

.0150 mol H2 x 22.4 L = 0.336 L of H2 (B) 1 mol

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Summary of Calculations of Molarity and Chemical Reactions

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