1. Pearl Harbor Day (1941) Mastering Chemistry Exam #3 -- Wednesday
Assignment #20 already posted
Due Sunday by 11:59PM
Molar Masses
Exam #3 -- Wednesday

2. The mole Only 3 options of what to do with the mole: mole particles
Avogadro’s #
mole
particles
molar mass
mole
mass
mole-to-mole ratio
mole “X”
mole “Y”

3. The mole Only 2 options of what to do with the mole:
2 places to find a mole-to-mole ratio:
a chemical formula
H2O
mole-to-mole ratio comes from the subscripts of the formula

4. The mole Only 2 options of what to do with the mole:
2 places to find a mole-to-mole ratio:
a chemical equation
3 H2(g) + 1 N2(g)  2 NH3(l)
mole-to-mole ratio comes from the coefficients of the balanced chemical equation.

5. ANY TIME you are going from an amount of one substance to an amount of another substance, you MUST use a mole to mole ratio.

6. ANY TIME you are going from an amount of one thing to an amount of another thing, work in 3 steps:
Step 1: convert the given information to moles
Step 2: use a mole-to-mole ratio
Step 3: convert the moles to whatever is asked for

7. mole-to-mole ratios
Silver nitrate is reacted with barium chloride:
How many grams of barium chloride are needed
to make 2.24 grams of solid?
2 AgNO3(aq) + BaCl2(aq)  2 AgCl(s) + Ba(NO3)2(aq)
game plan: g AgCl
game plan: g AgCl  mol AgCl
game plan:
game plan: g AgCl  mol AgCl  mol BaCl2
game plan: g AgCl  mol AgCl  mol BaCl2  g BaCl2
ANY TIME you are going from an amount of one substance to an amount of another substance, you MUST use a mole to mole ratio.
AgCl =
BaCl2 =

8. mole-to-mole ratios
Chromium (III) chlorate is reacted with sodium oxide:
a) What is the balanced chemical equation?
b) How many grams of sodium oxide are needed to react with grams of chromium (III) chlorate
Cr(ClO3)3 =
Na2O = g/mol

9. mole-to-mole ratios
Chromium (III) chlorate is reacted with sodium oxide:
2 Cr(ClO3)3(aq) + 3 Na2O(aq)  Cr2O3(s) + 6 NaClO3(aq)
b) How many grams of sodium oxide are needed to react with grams of chromium (III) chlorate
game plan: g Cr(ClO3)3  mol Cr(ClO3)3
game plan: g Cr(ClO3)3  mol Cr(ClO3)3  mol Na2O  g Na2O
game plan: g Cr(ClO3)3
game plan: g Cr(ClO3)3  mol Cr(ClO3)3  mol Na2O
game plan:

10. Mole – Mole revisited
What mass of liquid will be produced from the reaction of bismuth (V) hydroxide with 3.28 moles of sulfuric acid?
2 Bi(OH)5(s) + 5 H2SO4(aq)  Bi2(SO4)5(aq) + 10 H2O(l)
game plan:
game plan: mol H2SO4
game plan: mol H2SO4  mol H2O  g H2O
game plan: mol H2SO4  mol H2O

11. mole-to-mole ratios
What mass of liquid will be produced from the reaction of 203 g of bismuth (V) hydroxide with a sulfuric acid solution?
2 Bi(OH)5(s) + 5 H2SO4(aq)  Bi2(SO4)5(aq) + 10 H2O(l)
game plan: g Bi(OH)5
game plan:
game plan: g Bi(OH)5  mol Bi(OH)5
game plan: g Bi(OH)5  mol Bi(OH)5  mol H2O  g H2O
game plan: g Bi(OH)5  mol Bi(OH)5  mol H2O
Bi(OH)5 = g/mol

12. mole-to-mole ratios
2 Bi(OH)5(s) + 5 H2SO4(aq)  Bi2(SO4)5(aq) + 10 H2O(l)
Start with 3.28 mol H2SO4(aq)
Can make
118 grams of water
Start with 203 grams of Bi(OH)5
Can make
62.2 grams of water
What if we start with
3.28 mol H2SO4(aq)
AND
203 grams of Bi(OH)5?

13. ACME Tricycle Company
Tricycles are made of 1 frame, 2 pedals, and 3 wheels
A shipment comes in consisting of 1387 f, 2744 p, 4188 w.
How many tricycles can be built?

14. Limiting Reactants
Hydrogen gas and nitrogen gas combine to form ammonia gas (NH3). If you start with 18 moles of hydrogen and 11 moles of nitrogen, how many moles of ammonia can be made?
Hydrogen gas and nitrogen gas combine to form ammonia gas (NH3). If you start with 18 moles of hydrogen and 11 moles of nitrogen, how many moles of ammonia can be made?
3 H2(g) + N2(g)  2 NH3(g)
H2(g) + N2(g)  2 NH3(g)
H2(g) + N2(g)  NH3(g)
H2 is the limiting reactant (you run out of it first)
N2 is the excess reactant (you have more than you need)
Only 12 moles of NH3 can be made

15. 3 H2 + N2
2 NH3 +

16. theoretical yield of KC2H3O2
31.89 g of potassium sulfate is reacted with g of lead (II) acetate. What is the mass of potassium acetate can be formed?
K2SO4(aq) + Pb(C2H3O2)2(aq)  2 KC2H3O2(aq) + PbSO4(s)
g/mol
325.3 g/mol
98.15 g/mol
303.3 g/mol
limiting reactant
theoretical yield of KC2H3O2

17. 2 (NH4)3PO4(aq) + 3 CaS(aq)  3 (NH4)2S(aq) + Ca3(PO4)2(s)
g of ammonium phosphate is reacted with g of calcium sulfide. What is the mass of solid formed after the reaction is complete?
2 (NH4)3PO4(aq) + 3 CaS(aq)  3 (NH4)2S(aq) + Ca3(PO4)2(s)
g/mol
72.15 g/mol
68.17 g/mol
g/mol
Can make g Ca3(PO4)2
(NH4)3PO4 is the L.R. (0 left)