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Consider the following reaction.Cr2O3(s) + 3CCl4(l) 2CrCl3(s) + 3COCl2(g). 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?A) Cr2O3B) CCl4C) CrCl3D) COCl2E) there is no limiting reactant

Answer :

joseaaronlara

Answer:

The answer to your question is: letter B

Explanation:

Reaction

                 Cr2O3(s)   +   3CCl4(l)   ⇒  2CrCl3(s)  +   3COCl2(g)

From the information given and the reaction, we can conclude that:

Green solid = Cr2O3 (s)     "s" means solid

Colorless liquid = CCl4 (l)    "l" means liquid   and is the other reactant

Purple solid = CrCl3(s)        CrCl3 is purple and "s" solid

Then, as a green specks remains it means that the excess reactant is Cr2O3, so, CCl4 is the limiting reactant.

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