 # Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given (molarity) to convert.

## Presentation on theme: "Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given (molarity) to convert."— Presentation transcript:

Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given (molarity) to convert to moles, then use the molar ratio to convert to moles of the final product, then use concentration of the final product to convert to the final answer. Moles/L

Stoichiometry overview Recall that in stoichiometry the mole ratio provides a necessary conversion factor: grams (x)  moles (x)  moles (y)  grams (y) molar mass of xmolar mass of y mole ratio from balanced equation We can do something similar with solutions: volume (x)  moles (x)  moles (y)  volume (y) mol/L of xmol/L of y mole ratio from balanced equation

Calculate the volume of a 2.00 mol/L silver nitrate solution that is needed for 12.0 g of copper metal to react according to the following equation? 2 mol AgNO 3 1 mol Cu x # mol AgNO 3 = 12.0 g Cu 0.38 mol AgNO 3 = 1 mol Cu 63.5 g Cu x Cu(s) + 2AgNO 3 (aq)  Cu(NO 3 ) 2 + 2Ag(s) Calculate the moles of AgNO 3, then volume of AgNO 3 L = 0.38 mol AgNO 3 / 2.00 mol / L = 0.19L = 190mL

Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used? 1 mol H 2 SO 4 2 mol NH 3 x # mol H 2 SO 4 = 0.0244 L NH 3 0.02684 mol H 2 SO 4 = 2.20 mol NH 3 L NH 3 x H 2 SO 4 (aq) + 2NH 3 (aq)  (NH 4 ) 2 SO 4 (aq) Calculate mol H 2 SO 4, then mol/L = mol/0.0500 L mol/L = 0.02684 mol H 2 SO 4 / 0.0500 L = 0.537 mol/L

Calculate the volume of 0.0250 mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of 0.125 mol/L aluminum sulfate solution. 3 mol Ca(OH) 2 1 mol Al 2 (SO 4 ) 3 x # L Ca(OH) 2 = 0.0250 L Al 2 (SO 4 ) 3 = 0.375 L Ca(OH) 2 0.125 mol Al 2 (SO 4 ) 3 L Al 2 (SO 4 ) 3 x Al 2 (SO 4 ) 3 (aq) + 3Ca(OH) 2 (aq)  2Al(OH) 3 (s) + 3CaSO 4 (s) L Ca(OH) 2 0.0250 mol Ca(OH) 2 x

A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with a 0.250 mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed? 3 mol Na 2 CO 3 2 mol FeCl 3 x # L Na 2 CO 3 = 0.0750 L FeCl 3 = 0.0900 L Na 2 CO 3 = 90.0 mL Na 2 CO 3 0.200 mol FeCl 3 L FeCl 3 x 2FeCl 3 (aq) + 3Na 2 CO 3 (aq)  Fe 2 (CO 3 ) 3 (s) + 6NaCl(aq) L Na 2 CO 3 0.250 mol Na 2 CO 3 x

1.H 2 SO 4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H 2 SO 4 will be required to react completely with 75 mL of 0.50 mol/L NaOH? 2.How many moles of Fe(OH) 3 are produced when 85.0 L of iron(III) sulfate at a concentration of 0.600 mol/L reacts with excess NaOH? 3.What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution. Assignment

4.a) What volume of 0.20 mol/L AgNO 3 will be needed to react completely with 25.0 mL of 0.50 mol/L potassium phosphate? b) What mass of silver phosphate is produced from the above reaction? Assignment

Answers 2 mol Fe(OH) 3 1 mol Fe 2 (SO 4 ) 3 x # mol Fe(OH) 3 = 85 L Fe 2 (SO 4 ) 3 0.600 mol Fe 2 (SO 4 ) 3 L Fe 2 (SO 4 ) 3 x 1. H 2 SO 4 (aq) + 2NaOH(aq)  2H 2 O + Na 2 SO 4 (aq) = 102 mol 1 mol H 2 SO 4 2 mol NaOH x # L H 2 SO 4 = 0.075 L NaOH = 0.009375 L = 9.4 mL 0.50 mol NaOH L NaOH x 2. Fe 2 (SO 4 ) 3 (aq) + 6NaOH(aq)  2Fe(OH) 3 (s) + 3Na 2 SO 4 (aq) L H 2 SO 4 2.0 mol H 2 SO 4 x

3. 2NaOH (aq) + ZnCl 2 (aq)  Zn(OH) 2 (s) + 2NaCl (aq) 1 mol Zn(OH) 2 2 mol NaOH x # g Zn(OH) 2 = 0.0500 L NaOH = 6.21 g 2.50 mol NaOH L NaOH x 4a. 3AgNO 3 (aq) + K 3 PO 4 (aq)  Ag 3 PO 4 (s) + 3KNO 3 (aq) 99.40 g Zn(OH) 2 1 mol Zn(OH) 2 x 3 mol AgNO 3 1 mol K 3 PO 4 x # L AgNO 3 = 0.025 L K 3 PO 4 = 0.1875 L = 0.19 L 0.50 mol K 3 PO 4 L K 3 PO 4 x L AgNO 3 0.20 mol AgNO 3 x 1 mol Ag 3 PO 4 1 mol K 3 PO 4 x # g Ag 3 PO 4 = 0.025 L K 3 PO 4 = 5.2 g 0.50 mol K 3 PO 4 L K 3 PO 4 x 418.58 g Ag 3 PO 4 1 mol Ag 3 PO 4 x 4b. 3AgNO 3 (aq) + K 3 PO 4 (aq)  Ag 3 PO 4 (s) + 3KNO 3 (aq)

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