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
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