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 %

Potassium chloride

Iron (II) chloride

Copper (II) sulfate

Aluminum oxide

Iron (III) oxide

Remember to put roman numerals for the compounds where the metal has more than one possible oxidation state. For example, you see iron here with a +2 charge in one compound and then +3 charge in another. If the metal only has one possible oxidation state (like potassium and aluminum), you don't need a roman numeral.