Consider the following reaction: Cr₂O₃(s)+ 3CCl₄(l)→ 2CrCl₃(s)+ 3COCl₂(g)green colorless purple colorless solid liquid solid gas When the green solid is mixed with the colorless liquid, the mixture starts to bubble and fume. When all action has stopped, a dry purple solid containing solid green specks remains. Which substance is the limiting reactant?

Question

Quizplus · Accepted Answer

In order to determine the limiting reactant, we need to calculate the amount of product that can be formed with each reactant and compare them. Using the balanced chemical equation: 1 mol Cr₂O₃ produces 2 mol CrCl₃ 3 mol CCl₄ produces 3 mol COCl₂ First, we need to find the amount of product that can be formed with Cr₂O₃: - Since 1 mol of Cr₂O₃ is used and 2 mol of CrCl₃ are produced, the theoretical yield of CrCl₃ is 2/1 = 2 mol. Next, we need to find the amount of product that can be formed with CCl₄: - Since 3 mol of CCl₄ is used and 3 mol of COCl₂ are produced, the theoretical yield of COCl₂ is 3/3 = 1 mol. Therefore, CCl₄ is the limiting reactant because it produces less product than Cr₂O₃.

Consider the Following Reaction: Cr2O3(s)+ 3CCl4(l)→ 2CrCl3(s)+ 3COCl2(g)green Colorless Purple

Consider the Following Reaction: Cr₂O₃(s)+ 3CCl₄(l)→ 2CrCl₃(s)+ 3COCl₂(g)green Colorless Purple