Answer:
52.6 g C₆H₆O₂
Explanation:
Unfortunately, the problem does not specify which of the reactants is the limiting reagent. This means you must convert both starting amounts into the product to identify the final answer.
To find the mass of hydroquinone (C₆H₆O₂) produced, you need to (1) calculate the molar masses of the reactants and product, then (2) convert grams of the reactants to grams of the product (via the molar masses and the mole-to-mole ratio from equation coefficients), and then (3) determine the limiting reagent and the actual answer.
(Step 1)
Molar Mass (C₆H₅OH): 6(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol
Molar Mass (C₆H₅OH): 94.112 g/mol
Molar Mass (H₂O₂): 2(1.008 g/mol) + 2(15.998 g/mol)
Molar Mass (H₂O₂): 34.012 g/mol
Molar Mass (C₆H₆O₂): 6(12.011 g/mol) + 6(1.008 g/mol) + 2(15.998 g/mol)
Molar Mass (C₆H₆O₂): 110.11 g/mol
(Step 2)
It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs of the given values.
45.0 g C₆H₅OH 1 g 1 mole C₆H₆O₂ 110.11 g
------------------------ x -------------- x ------------------------- x ----------------- =
94.112 g 1 mole C₆H₅OH 1 mole
= 52.6 g C₆H₆O₂
20.0 g H₂O₂ 1 mole 1 mole C₆H₆O₂ 110.11 g
------------------- x ----------------- x ------------------------- x --------------- =
34.012 g 1 mole H₂O₂ 1 mole
= 64.7 g H₂O₂
(Step 3)
The limiting reagent of the reaction is C₆H₅OH because it is not able to produce as much product as H₂O₂. As such, product stops being made as C₆H₅OH runs out, making the final amount 52.6 g C₆H₆O₂.
