Science Chemistry: Matter and Change
Science Chemistry: Matter and Change Glencoe Online
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 Problem of the Week

Solutions

Teachers Notes:

Problem 1:
C6H4(OH)2 (l) + 3 H2O2 (l) → C6H4(OH)2 (l) + 4 H2O (l) + O2 (g)

Problem 2:
1.00g H2O2 x (1 mol H2O2 / 34.01 g H2O2 ) = 0.0294 mol H2O2

2.00 g C6H4(OH)2 x (1mol C6H4(OH)2 /110.11 g C6H4(OH)2) = 0.0182 mol C6H4(OH)2
0.0294 mol H2O2 /0.0182 mol C6H4(OH)2 = 1.62 mol H2O2 / 1 mol C6H4(OH)2

According to the equation there should be 3 mol H2O2 /1 mole C6H4(OH)2, Therefore the hydrogen peroxide in the limiting reactant and the hydroquinone is the excess reactant.

1.00g H2O2 x (1 mol H2O2/ 34.01 g H2O2) x (1 mol C6H4(OH)2/3 mol H2O2 ) x (110.11 g C6H4(OH)2/ 1mol C6H4(OH)2) = 1.08 g C6H4(OH)2 are needed.
2.00 g - 1.08 g = 0.92 g C6H4(OH)2 in excess

Problem 3:
1.00g H2O2 x (1 mol H2O2/ 34.01 g H2O2) x (4 mol H2O/3 mol H2O2) x (18.1 g H2O/1 mol H2O) = 0.706 g H2O are produced.
 
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Matter and Change