Question

I need help with the solution

Answer 1

See below

Step-by-step explanation:

Mole weight of C6H5OH = 94 gm/mole (approx)

45 gm / 94 gm/mole = .479 Moles of phenol

Mole wt of H2O2 = 34 (approx)

20 gm / 34 gm /mole = .588 moles of hydrogen peroxide

From the equation.....it takes equal moles of these to react to produce the same number of moles of hydroquinone.....

there isn't enough phenol to match the peroxide....so phenol is limiting reactant and you will only get .479 moles of hydroquinone

hydroquinone mole weight = 110 gm / mole (approx)

.479 mole * 110 gm/mole = 52.69 gm produced

Answer 2

52.6 g C₆H₆O₂

Step-by-step 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.