Consider the following reaction: Cr₂O₃(s) + 3CCl₄(l) $\rightarrow$ 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

To determine the limiting reactant, we need to calculate how much of each reactant is needed to completely react with each other. We can do this by using stoichiometry and comparing the amounts of each reactant we have to the amounts required by the balanced chemical equation. Starting with Cr₂O₃: 1 mole of Cr₂O₃ reacts with 3 moles of CCl₄. We have the green solid which contains both Cr₂O₃ and CrCl₃, but we don't know the exact amount of Cr₂O₃ present. Therefore, we cannot determine if this is the limiting reactant. Moving on to CCl₄: 1 mole of CCl₄ reacts with 1 mole of Cr₂O₃, and we have 3 moles of CCl₄. Since we have more than enough CCl₄ to react with the Cr₂O₃ present, CCl₄ cannot be the limiting reactant. Moving on to CrCl₃: For every 1 mole of Cr₂O₃, 2 moles of CrCl₃ are formed. We don't know the exact amount of Cr₂O₃ present, but we do know that the amount of CrCl₃ formed will be twice that amount. Since the purple solid that remains contains solid green specks, we know that some Cr₂O₃ did not react and is still present in the mixture. Therefore, the CrCl₃ formed must be less than or equal to the amount of unreacted Cr₂O₃ present. This means that CrCl₃ is the limiting reactant. Therefore, the answer is B: CCl₄.

Consider the Following Reaction: Cr2O3(s) + 3CCl4(l) →\rightarrow→ 2CrCl3(s) + 3COCl2(g) Green Colorless Purple Colorless Solid Liquid Solid

Consider the Following Reaction: Cr₂O₃(s) + 3CCl₄(l) $\rightarrow$ 2CrCl₃(s) + 3COCl₂(g) Green Colorless Purple Colorless Solid Liquid Solid