# Question #c8eb0

Nov 18, 2016

You have been given a series of stoichiometric equations.....

#### Explanation:

For $a .$ $2 A l + 3 C {l}_{2} \rightarrow 2 A l C {l}_{3}$

Two equivalents, 2 moles if you like, of aluminum metal react with 3 equiv of dichlorine gas to give 2 equiv aluminum trichloride.

You starting conditions propose that you have $3.6$ mol of aluminum metal, and $5.3$ mol of $C {l}_{2}$ gas. Clearly (?), there is insufficient chlorine gas for all the metal to react (given the stoichiometry, 3.6 mol of metal require $5.4$ mol of bimolecular chlorine gas, and you only have $5.3$ mol of $C {l}_{2}$ gas.

And thus here, $C {l}_{2}$ gas is the limiting reagent, which governs the extent of the reaction. At most $5.3 \times \frac{2}{3} \cdot m o l$ aluminum chloride can be generated given the limiting quantity of chlorine.

I am reluctant to do the remaining problems, because I think you should work thru them yourself. You have been given balanced chemical equations that describe the equivalence, the stoichiometry. Make an attempt, and someone here will help you.