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dolphi86 [110]
3 years ago
12

Consider the following reaction.Cr2O3(s) + 3CCl4(l) 2CrCl3(s) + 3COCl2(g). When the green solid is mixed with the colorless liq

uid, 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
Chemistry
1 answer:
Elena L [17]3 years ago
6 0

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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A gas sample occupies a volume of 1.264 L when the temperature is 168.0 °C and the pressure is 946.6 torr. How many molecules ar
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Answer:

0.26×10²³ molecules

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Pressure = 946.6 torr (946.6/760 = 1.25 atm)

Now we will determine the number of moles.

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n = 1.25  atm ×1.264 L / 0.0821 atm.L/ mol.K   ×441 K

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12.2 C
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