The stoichiometry calculations done in Chapter 8 involved pure substances and samples whose mass could be determined by weighing. In Chapter 9, the reactant was often a gas. In these cases the mass of the sample could be determined if we knew its pressure, volume, and temperature by applying the ideal gas equation and the concept of molar volume. Now, by using the concentration of a solution as a conversion factor, we can extend stoichiometric calculations to include solutions.

Example: What weight of barium sulfate is precipitated by the addition of an excess of sulfuric acid to 55.6 mL of 0.54 M barium chloride? Solution Equation BaCl2 + H2SO4 BaSO4 + 2 HCl Given 55.6 mL of 0.54 M BaCl2 Conversion factors 0.54 mol BaCl2 in 1 L of 0.54 M BaCl2 0.54 mmol BaCl2 in 1 mL of 0.54 M BaCl2 1 mol BaCl2 forms 1 mol BaSO4 1 mol BaSO4 weighs 233.4 g Arithmetic equation Answer 7.0 g BaSO4

In the next example, two solutions are used. The concentration and volume of the first are known; only the concentration of the second is known. Its volume can be calculated.

Example: What volume of 0.154 M sodium hydroxide will completely react with 25.0 mL of 0.0952 M hydrochloric acid? Solution Equation NaOH + HCl NaCl + H2O Wanted ? ml of 0.154 M NaOH Given 25.0 mL of 0.0952 M HCl Conversion factors 1 mL of 0.0952 M HCl contains 0.0952 mmol HCl 1 mol NaOH reacts with 1 mol HCl 1 mL of 0.154 M NaOH contains 0.154 mmol NaOH Arithmetic equation Answer 15.5 mL of 0.154 M NaOH

The next example illustrates the use of molarity in stoichiometric problems invloving gases.