Solutions stoichiometry.

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Solutions stoichiometry.

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Solutions stoichiometry

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 x molar mass of y mole ratio from balanced equation We can do something similar with solutions: volume (x)  moles (x)  moles (y)  volume (y) C (mol/L) of x C (mol/L) of y mole ratio from balanced equation

H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) 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? H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) c = ? c = 2.20 mol/L v = 0.05 L v = 0.0244L n = ? n = ? (2) (1b) (1a) 2.20 mol NH3 L NH3 x 1 mol H2SO4 2 mol NH3 x (n) H2SO4= 0.0244 L NH3 0.02684 mol H2SO4 = (2) Then do c=n/v for H2SO4 n = ( V x c ) x mole ratio (1a) (1b)

Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) 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. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) # L Ca(OH)2= 0.0250 L Al2(SO4)3 0.125 mol Al2(SO4)3 L Al2(SO4)3 x 3 mol Ca(OH)2 1 mol Al2(SO4)3 x L Ca(OH)2 0.0250 mol Ca(OH)2 x = 0.375 L Ca(OH)2

2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) 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? 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) 4 step problem! (1st being writing the balanced chemical equation) V (FeCl3) n (FeCl3) n (Na2CO3)  V (Na2CO3) C (FeCl3) C (Na2CO3) mole ratio from balanced equation

2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) Answers 1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Calculate mol H2SO4, then mol/L = mol/0.0500 L # mol H2SO4= 2.20 mol NH3 L NH3 x 1 mol H2SO4 2 mol NH3 x 0.0244 L NH3 0.02684 mol H2SO4 = mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L 2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) # L Ca(OH)2= 0.0250 L Al2(SO4)3 0.125 mol Al2(SO4)3 L Al2(SO4)3 x 3 mol Ca(OH)2 1 mol Al2(SO4)3 x L Ca(OH)2 0.0250 mol Ca(OH)2 x = 0.375 L Ca(OH)2

3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) Answers 3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) # L Na2CO3= 0.0750 L FeCl3 0.200 mol FeCl3 L FeCl3 x 3 mol Na2CO3 2 mol FeCl3 x L Na2CO3 0.250 mol Na2CO3 x = 0.0900 L Na2CO3 = 90.0 mL Na2CO3