# Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s) +...

## Question:

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

{eq}\rm 2Al(s) + 3Cl_2(g) \to 2AlCl_3(s) {/eq}

You are given 14.0 g of aluminum and 19.0 g of chlorine gas.

1) If you had excess chlorine, how many moles of aluminum chloride could be produced from 14.0 g of aluminum?

2) If you had excess aluminum, how many moles of aluminum chloride could be produced from 19.0 g of chlorine gas, {eq}\rm Cl_2 {/eq}?

## Limiting Reactant

In a chemical reaction, the maximum amount of product that can be produced is determined by the initial amount of the limiting reactant. The limiting reactant is the reactant present in smaller amounts compared to the excess reactant according to the chemical equivalence in the balanced reaction.

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### Part 1)

If chlorine is in excess, then, aluminum is the limiting reactant. Let us calculate the initial moles of aluminum (molar mass = 26.98 g/mol).

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