Chemistry Review Manual 2

Solutions:
1.1 First, write a balanced reaction:
2 Ca(s) + O 2 (g) → 2 CaO(s)
Initial amount of calcium:
( Ca)
( Ca)
m
4.20 g
n( Ca ) = = = 0.104
8
mol
M 40.08 g / mol
Initial amount of oxygen:
( O2
)
( O )
m
1.60 g
n( O
2 ) = = = 0.0500 mol
M 31.9988 g / mol
2
Since O 2 and Ca react in a 1:2 ratio, O 2 is the limiting reactant – there is less than 0.0524
mol (the amount needed to consume the entire 0.104 8 mol of Ca).
Amount of CaO produced = 2 × (amount of O 2 ) = 2 x 0.500 mol = 0.100 mol
Mass of CaO produced = (0.100 mol) × (56.08 g/mol) = 5.61 g
1.2 (a) Because the density is expressed in g mL -1 , we must convert m 3 to mL. Since 1 m =
10 2 cm,
2500 m 3 = 2500 × (10 2 cm) 3 = 2500 × 10 6 cm 3 = 2.500 × 10 9 mL
Mass of water = volume × density = 2.500 × 10 9 mL × 1.00 g / mL = 2.500 × 10 9 g
Amount of water =
( )
( HO)
( H O)
m 2.500×
10 g
n H O = 1.387 10 mol
9
2 8
2
= =
7×
M
2
18.015 g / mol
Multiply by the Avogadro constant to get the number of water molecules:
1.387 7 ×10 8 mol × 6.022 ×10 23 molecules mol −1 = 8.356 7 ×10 31 molecules
(b) The total volume of 8.356 7 ×10 31 cubes is
8.356 7 ×10 31 cubes × 1.00 cm 3 /cube = 8.356 7 ×10 31 cm 3
= 8.356 7 ×10 31 cm 3 (10 −5 km/cm) 3 = 8.357 ×10 16 km 3
This volume corresponds to giant cube more than half of 1 million kilometers along each
side.
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