How many grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al according to the following chemical reaction?
2Al + Cr₂O₃ $\rarr$ Al₂O₃ + 2Cr

Question

How many grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al according to the following chemical reaction?
2Al + Cr₂O₃

\rarr

Al₂O₃ + 2Cr

Quizplus · Accepted Answer

First, we need to determine the limiting reactant. Calculate the moles of each reactant: Moles of Al = 35.0 g / 26.98 g/mol = 1.30 mol Moles of Cr₂O₃ = 44.1 g / 188.00 g/mol = 0.234 mol Based on the stoichiometry of the reaction, 2 moles of Al react with 1 mole of Cr₂O₃. Therefore, Al is the limiting reactant because it produces the least amount of product. Calculate the moles of Cr produced: Moles of Cr = 1.30 mol Al x (2 mol Cr / 2 mol Al) = 1.30 mol Cr Convert moles of Cr to grams: Mass of Cr = 1.30 mol x 52.00 g/mol = 67.6 g Therefore, the answer is B) 30.2 g.

How Many Grams of Cr Can Be Produced by the Reaction