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Practice with Molar Mass Problems

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Practice with Molar Mass Problems

x 1 mol H2

2.02 g H2

How many moles of H2 are in 100 g of H2?

# mol H2

= 100 g H2

= 49.5 mol H2

x 1 mol CuSO4

159.61 g CuSO4

300 g of CuSO4 is needed in an experiment. How many moles does this represent?

# mol CuSO4 =

300 g CuSO4

= 1.88 mol CuSO4

x 143.32 g AgCl

1 mol AgCl

A chemical reaction requires 23.78 moles of silver chloride. How many grams is this?

# g AgCl =

23.78 mol AgCl

= 3408 g AgCl

x 6.02x1023 molecules

1 mol H2O

x 1 mol H2O

18.02 g H2O

How many molecules are in 73 grams H2O?

# H2O molecules =

73 g H2O

= 2.4 x 1024 molecules H2O

x 1 mol Ca3(PO4)2

310.18 g Ca3(PO4)2

255 g of calcium phosphate are produced in a chemical reaction. How many moles of calcium phosphate does this represent?

# mol Ca3(PO4)2 =

255 g Ca3(PO4)2

= 0.822 mol Ca3(PO4)2

2 mol H2O

18.01 g H2O

265 g H2O

x

x

=

1 mol O2

1 mol H2O

According to the equation 2H2 + O2 2H2O, how many grams of H2O would be produced if 7.35 mol of O2 is used up? (hint: you will need two conversion factors – 1 from the balanced equation and 1 from a molar mass)

# g H2O=

7.35 mol O2

Complete the following chart (#7 - #13):

Formula

Molar mass

(g/mol)

Mass

(g)

Moles

(mol)

Complete the following chart (answers):

FeSO4

151.9

500

3.29

(NH4)2CO3

96.1

192.2

2

SnO2

150.7

50

0.332

Sb2O5

323.6

80.9

0.25

NaClO4

122.4

100

0.817

Mg(IO3)2

374.1

1196.8

3.2

CoCl2.H2O

147.8

332

2.246

14. AgCl = 143.35 g/mol

#g = 2 mol x 143.35 g/mol =286.7 g (2)

15. H2 = 2.016 g/mol

#mol = 100 g x mol/2.016 g =49.6 mol (2)

16. CuSO4 = 159.62 g/mol

#mol= 300 g x mol/159.62 g=1.879 mol (2)

17. KClO = 90.55 g/mol

#mol = 250 g x mol/90.55 g =2.76 mol (2)

x 6.02x1023 molecules

1 mol H2O

x 1 mol H2O

18.02 g H2O

# H2O molecules =

73 g H2O

= 2.4 x 1024 molecules H2O

x 253.80 g I2

1 mol I2

x 3 mol I2

2 mol Al

Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum.

Al + I2 AlI3

2Al + 3 I2 2 AlI3

0.50 mol Al

= 190 g I2

x 253.80 g I2

1 mol I2

x 3 mol I2

x 1 mol Al

2 mol Al

26.98 g Al

Calculate the mass in grams of iodine required to react completely with 0.50 g of aluminum.

Al + I2 AlI3

2Al + 3 I2 2 AlI3

0.50 g Al

= 7.1 g I2