The product of the reaction between Aluminium Oxide (Al2O3) and Carbon (C) is Aluminium (Al) and carbon monoxide (CO). The reaction of 60.0 g of Al2O3 with 30.0 g of C produced 22.5 g of Al. What is the percent yield for this reaction?

Percent yield = 70.8%

Explanation:

Equation of the reaction:

Al2O3 + 3C > 2Al + 3CO

From the reaction equation, 1 mole of Aluminum oxide produces 2 moles of Aluminum when reacted with 3 moles of carbon.

Molar mass of Aluminum oxide = 102 g; molar mass of Aluminum = 27 g; molar mass of C = 12

3 moles of C = 36 g

2 moles Al = 54 g

102 g of Al2O3 reacts with 36 g of C

60 g of Al2O3 will react with (36/102) × 60 g of C = 21.2 g of C

Therefore, Al2O3 is the limiting reactant while C is in excess.

102 g of Al2O3 produces 54 g of Al

60 g of Al2O3 will produce (54/102) × 60 of Al = 31.8 g of Al

Percent yield = (actual yield/ theoretical yield) × 100%

Percent yield = (22.5 g / 31.8 g) × 100%

Percent yield = 70.8 %

i believe the answer is a.