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Solutions. stoichiometry. molar mass of x. molar mass of y . mol/L of x. mol/L of y . mole ratio from balanced equation . mole ratio from balanced equation . Stoichiometry overview. Recall that in stoichiometry the mole ratio provides a necessary conversion factor:

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slide1

Solutions

stoichiometry

stoichiometry overview

molar mass of x

molar mass of y

mol/L of x

mol/L of y

mole ratio from balanced equation

mole ratio from balanced equation

Stoichiometry overview
  • Recall that in stoichiometry the mole ratio provides a necessary conversion factor:

grams (x)  moles (x)  moles (y)  grams (y)

  • We can do something similar with solutions:
  • volume (x)  moles (x)  moles (y)  volume (y)
  • Read pg. 351-353. Try Q 1-3a.
pg 353 question 1

2.20 mol NH3

1 mol H2SO4

x

x

L NH3

2 mol NH3

0.02684 mol H2SO4

=

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?

H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)

Calculate mol H2SO4, then mol/L = mol/0.0500 L

# mol H2SO4=

0.0244 L NH3

mol/L = 0.02684 mol H2SO4/ 0.0500 L = 0.537 mol/L

pg 353 question 2

0.125 mol Al2(SO4)3

3 mol Ca(OH)2

L Ca(OH)2

x

x

x

L Al2(SO4)3

0.0250 mol Ca(OH)2

1 mol Al2(SO4)3

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.

Al2(SO4)3(aq)+3Ca(OH)2(aq)2Al(OH)3(s)+3CaSO4(s)

# L Ca(OH)2=

0.0250 LAl2(SO4)3

= 0.375 L Ca(OH)2

pg 353 question 3

0.200 mol FeCl3

L Na2CO3

3 mol Na2CO3

x

x

x

L FeCl3

0.250 mol Na2CO3

2 mol FeCl3

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?

2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)

# L Na2CO3=

0.0750 LFeCl3

= 0.0900 L Na2CO3 = 90.0 mL Na2CO3

answers

0.125 mol Al2(SO4)3

2.20 mol NH3

3 mol Ca(OH)2

L Ca(OH)2

1 mol H2SO4

x

x

x

x

x

L Al2(SO4)3

L NH3

0.0250 mol Ca(OH)2

1 mol Al2(SO4)3

2 mol NH3

0.02684 mol H2SO4

=

Answers

# mol H2SO4=

1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)

Calculate mol H2SO4, then mol/L = mol/0.0500 L

0.0244 L NH3

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 LAl2(SO4)3

= 0.375 L Ca(OH)2

answers7

0.200 mol FeCl3

L Na2CO3

3 mol Na2CO3

x

x

x

L FeCl3

0.250 mol Na2CO3

2 mol FeCl3

3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)

Answers

# L Na2CO3=

0.0750 LFeCl3

= 0.0900 L Na2CO3 = 90.0 mL Na2CO3

assignment
Assignment
  • H2SO4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H2SO4 will be required to react completely with 75 mL of 0.50 mol/L NaOH?
  • 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?
  • 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.
assignment9
Assignment
  • a) What volume of 0.20 mol/L AgNO3 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?

  • 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?
answers10

0.50 mol NaOH

0.600 mol Fe2(SO4)3

L H2SO4

1 mol H2SO4

2 mol Fe(OH)3

x

x

x

x

x

L NaOH

L Fe2(SO4)3

2.0 mol H2SO4

1 mol Fe2(SO4)3

2 mol NaOH

Answers

# L H2SO4=

0.075 LNaOH

1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq)

= 0.009375 L = 9.4 mL

2. Fe2(SO4)3(aq)+6NaOH(aq)2Fe(OH)3(s)+3Na2SO4(aq)

# mol Fe(OH)3=

85 L Fe2(SO4)3

= 102 mol

slide11

0.50 mol K3PO4

0.50 mol K3PO4

2.50 mol NaOH

1 mol Zn(OH)2

1 mol Ag3PO4

3 mol AgNO3

418.58 g Ag3PO4

L AgNO3

99.40 g Zn(OH)2

x

x

x

x

x

x

x

x

x

L K3PO4

L K3PO4

L NaOH

1 mol Zn(OH)2

1 mol Ag3PO4

0.20 mol AgNO3

1 mol K3PO4

1 mol K3PO4

2 mol NaOH

# g Zn(OH)2=

= 6.21 g

0.0500 LNaOH

3. 2NaOH(aq)+ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq)

4a. 3AgNO3(aq)+K3PO4(aq)Ag3PO4(s) +3KNO3(aq)

# L AgNO3 =

= 0.1875 L = 0.19 L

0.025 LK3PO4

4b. 3AgNO3(aq)+K3PO4(aq)Ag3PO4(s) +3KNO3(aq)

# g Ag3PO4=

= 5.2 g

0.025 LK3PO4

slide12

1 mol Al(OH)3

1 mol Al(OH)3

x

x

1 mol Al(NO3)3

3 mol NaOH

=

=

21.4 g Al(OH)3

9.36 g Al(OH)3

1.50 mol NaOH

0.500 mol Al(NO3)3

77.98 g Al(OH)3

77.98 g Al(OH)3

x

x

x

x

L NaOH

L Al(NO3)3

1 mol Al(OH)3

1 mol Al(OH)3

# g Al(OH)3=

0.550 LAl(NO3)3

5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)

# g Al(OH)3=

0.240 L NaOH

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