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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.