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

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.

the answer is b & c i believe : )