If the reaction of 23.3 grams of Cr2O3 produces 5.35 grams
of Al2O3, what is the percent yield?

34.2%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Cr₂O₃ + 2Al —> Al₂O₃ + 2Cr

Next, we shall determine the mass of Cr₂O₃ that reacted and the mass of Al₂O₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Cr₂O₃ = (52×2) + (3×16)

= 104 + 48

= 152 g/mol

Mass of Cr₂O₃ from the balanced equation = 152 × 1 = 152 g

Molar mass of Al₂O₃ = (27×2) + (16×3)

= 54 + 48

= 102 g/mol

Mass of Al₂O₃ from the balanced equation = 1 × 102 = 102 g

SUMMARY:

From the balanced equation above,

152 g of Cr₂O₃ reacted to produce 102 g of Al₂O₃.

Next, we shall determine the theoretical yield of Al₂O₃. This can be obtained as follow:

From the balanced equation above,

152 g of Cr₂O₃ reacted to produce 102 g of Al₂O₃.

Therefore, 23.3 g of Cr₂O₃ will react to produce = (23.3 × 102)/152 = 15.64 g of Al₂O₃.

Thus, the theoretical yield of Al₂O₃ is 15.64 g.

Finally, we shall determine the percentage yield of Al₂O₃. This can be obtained as follow:

Actual yield of Al₂O₃ = 5.35 g

Theoretical yield of Al₂O₃ = 15.64 g.

Percentage yield of Al₂O₃ =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.35 / 15.64 × 100

Percentage yield of Al₂O₃ = 34.2%

the answer is true.

it's filtration, but this is an assumption so my apologies if i got this