Aluminum reacts with chlorine gas to produce aluminum chloride crystals. How many grams of aluminum chloride will be produced if the reaction begins with 135.0g of Al and 475.0 g chlorine gas.

596 g of AlCl₃

Explanation:

The reaction is:

2Al + 3Cl₂  →  2AlCl₃

Firstly we need to determine the limiting reactant and we need to determine the moles of each

135 g . 1 mol / 26.98 g = 5 moles of Al

475 g . 1mol / 70.9 g = 6.7 moles of chlorine

Ratio in the reaction is 3:2

3 moles of chlorine gas need 2 moles of Al to react

Then, our 6.7 moles of gas might react to (6.7 . 2 ) /3 = 4.47 moles of Al

We have 5 moles of Al and we need 4.47 moles, so this is the excess reactant. In conclussion, Cl₂ is the limiting.

2 moles of Al react to 3 moles of gas

Our 5 moles of Al may react to (5 . 3 )/2 = 7.5 moles of gas

It's ok because we only have 6.7 moles, and there is not enough gas.

Now we calculate the product's mass

3 moles of gas can produce 2 moles of salt

Then, our 6.7 moles of gas will produce (6.7 . 2) /3 = 4.47 moles

We convert moles to mass: 4.47 mol . 133.33 g/mol = 596 g of AlCl₃

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