Solution Stoichiometry

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Presentation transcript:

Solution Stoichiometry

Solution Stoichiometry The major difference compared to gravimetric and gas stoichiometry is that we use molarity (mol/L) as a conversion factor rather than molar mass or molar volume Solution stoichiometry – the procedure for calculating the molar concentration or volume of solution products or reactants

Solution Stoichiometry Steps Write a balanced chemical equation. Convert the given measurement to moles using molarity as a conversion factor. Use the mole ratio from the balanced chemical equation to calculate moles of unknown. Convert the moles of unknown to the final quantity requested using the appropriate conversion factor.

Example #1 Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer. What volume of 14.8mol/L NH3(aq) is needed to react with 1000 L of 12.9mol/L of H3PO4(aq)? 2NH3(aq) + H3PO4(aq)  (NH4)2HPO4(aq) V = ? V = 1000 L M = 14.8mol/L M = 12.9 mol/L 1000 L H3PO4 x 12.9 mol H3PO4 x 1 L H3PO4 2 mol NH3 x 1 mol H3PO4 1 L NH3 = 14.8 mol NH3 1.74 x 103 L

Example #2 In an experiment, a 10.00 mL sample of sulfuric acid solution reacts completely with 15.9 mL of 0.150 mol/L potassium hydroxide. Calculate the molarity of the sulfuric acid. H2SO4(aq) + 2KOH(aq)  2H2O(l) + K2SO4(aq) V = 10.00 mL V = 15.9 mL M = ? M = 0.150 mol/L 0.0159 L KOH x 0.150 mol KOH x 1 L KOH 1 mol H2SO4 = 2 mol KOH 0.00119 mol H2SO4 M = mol = L 0.00119 mol H2SO4 = 0.0100 L H2SO4 0.119 M H2SO4