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Stoichiometry overview

Recall that in stoichiometry the mole ratio provides a necessary conversion factor: molar mass of x molar mass of y grams (x)  moles (x)  moles (y)  grams (y) mole ratio from balanced equation • We can do something similar with solutions: mol/L of x mol/L of y volume (x)  moles (x)  moles (y)  volume (y) • mole ratio from balanced equation Read pg. 351-353. Try Q 1-3a.

Pg. 353, Question 1

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?

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 H 2 SO 4 = 0.0244 L NH 3 x 2.20 mol NH 3 L NH 3 mol/L = 0.02684 mol H 2 SO 4 x 1 mol H 2 SO 4 2 mol NH 3 = 0.02684 mol H 2 SO 4 / 0.0500 L = 0.537 mol/L

Pg. 353, Question 2

Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. 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.

Al 2 (SO 4 ) 3 (aq) + 3Ca(OH) 2 (aq)



# L Ca(OH) 2 = 0.0250 L Al 2 (SO 4 ) x 3 0.125 mol Al 2 (SO 4 ) 3 L Al 2 (SO 4 ) 3 x 3 mol Ca(OH) 2 1 mol Al 2 (SO 4 ) 3 = 0.375 L Ca(OH) 2 2Al(OH) 3 (s) + 3CaSO 4 (s) x L Ca(OH) 0.0250 mol Ca(OH) 2 2

Pg. 353, Question 3

A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with an excess of 0.250 mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed?

2FeCl 3 (aq) + 3Na 2 CO 3 (aq)



Fe 2 (CO 3 ) 3 (s) + 6NaCl(aq) # L Na 2 CO 3 = 0.0750 L FeCl 3 x 0.200 mol FeCl L FeCl 3 = 0.0900 L Na 2 CO 3 3 x 3 mol Na 2 CO 3 2 mol FeCl 3 = 90.0 mL Na 2 CO 3 x L Na Na 2 2 CO CO 3 3 0.250 mol

Answers

1. 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 H 2 SO 4 = 0.0244 L NH 3 x 2.20 mol NH 3 L NH 3 mol/L = 0.02684 mol H 2 SO 4 x 1 mol H 2 SO 4 2 mol NH 3 = 0.02684 mol H 2 SO 4 / 0.0500 L = 0.537 mol/L 2. Al 2 (SO 4 ) 3 (aq) + 3Ca(OH) 2 (aq) # L Ca(OH) 2 = 0.0250 L Al 2 (SO 4 ) x 3 0.125 mol Al 2 (SO 4 ) 3 L Al 2 (SO 4 ) 3 x 3 mol Ca(OH) 2 1 mol Al 2 (SO 4 ) 3 = 0.375 L Ca(OH) 2



2Al(OH) 3 (s) + 3CaSO 4 (s) x L Ca(OH) 2 0.0250 mol Ca(OH) 2

Answers

3. 2FeCl 3 (aq) + 3Na 2 CO 3 (aq)



Fe 2 (CO 3 ) 3 (s) + 6NaCl(aq) # L Na 2 CO 3 = 0.0750 L FeCl 3 x 0.200 mol FeCl L FeCl 3 = 0.0900 L Na 2 CO 3 3 x 3 mol Na 2 CO 3 2 mol FeCl 3 = 90.0 mL Na 2 CO 3 x L Na Na 2 2 CO CO 3 3 0.250 mol

Assignment

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 precipitate is produced from the above reaction?

5. What mass of precipitate should result when 0.550 L of 0.500 mol/L aluminum nitrate solution is mixed with 0.240 L of 1.50 mol/L sodium hydroxide solution?

Answers

1. H 2 SO 4 (aq) + 2NaOH(aq)  2H 2 O + Na 2 SO 4 (aq)

# L H 2 SO 4 = 0.075 L NaOH L NaOH 1 mol H 2 SO 4 2 mol NaOH x L H 2 SO 4 2.0 mol H 2 SO 4 = 0.009375 L = 9.4 mL 2. Fe 2 (SO 4 ) 3 (aq) + 6NaOH(aq)



2Fe(OH) 3 (s) + 3Na 2 SO 4 (aq) # mol Fe(OH) 3 = 85 L Fe 2 (SO 4 ) 3 x 0.600 mol Fe 2 (SO 4 ) 3 L Fe 2 (SO 4 ) 3 x 2 mol Fe(OH) 3 1 mol Fe 2 (SO 4 ) 3 = 102 mol

3. 2NaOH

(aq) # g Zn(OH) 2 = 0.0500 L NaOH

+ ZnCl 2

(aq)

 Zn(OH) 2

(s) x 2.50 mol NaOH L NaOH 1 mol Zn(OH) 2 mol NaOH 2 x

+ 2NaCl

(aq) = 6.21 g 99.40 g Zn(OH) 2 1 mol Zn(OH) 2 4a. 3AgNO 3 (aq) + K 3 PO 4 (aq) # L AgNO 3 =



Ag 3 PO 4 (s) + 3KNO 3 (aq) = 0.1875 L = 0.19 L 0.025 L K 3 PO 4 x L K 3 PO 4 # g Ag 3 PO 4 = 3 PO 4 x 3 mol AgNO 3 1 mol K 3 PO 4 4b. 3AgNO 3 (aq) + K 3 PO 4 (aq)



x L AgNO 3 0.20 mol AgNO Ag 3 PO 4 (s) + 3KNO 3 (aq) = 5.2 g 3 0.025 L K 3 PO 4 x 0.50 mol K 3 PO 4 L K 3 PO 4 x 3 PO 1 mol K 3 PO 4 4 x 418.58 g Ag 3 PO 4 1 mol Ag 3 PO 4

5. Al(NO 3 ) 3 (aq) + 3NaOH(aq)



Al(OH) 3 (s) + 3NaNO 3 (aq) # g Al(OH) LAl(NO 3 ) x 3 3 = L Al(NO 3 ) 3 3 ) 3 x 1 mol Al(OH) 3 1 mol Al(NO 3 ) 3 = x 21.4 g Al(OH) 3 3 3 # g Al(OH) L NaOH x 3 = L NaOH x 1 mol Al(OH) 3 3 mol NaOH = x 1 mol Al(OH) 3 9.36 g Al(OH) 3 3

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