The following data was collected when a reaction was performed experimentally in the laboratory.

Reaction Data

Reactants Products

Al (NO3) 3 NaCl NaNO3 AlCl3

Starting Amount in Reactio 4 moles 9 moles??

Determine the maximum amount of AlCl3 that was produced during the experiment. Explain how you determined this amount.

Answer: 764.91 grams of sodium nitrate will be formed.

Explanation:

Stoichiometric ratio of alluminium nirtate and sodium chloride is 1 : 3

According to the reaction one mole of aluminium nitrate reacts with 3 mole of sodium chloride.

Then 4 moles of aluminum nitrate will require : 12 mole of NaCl

According to question starting amount taken was

4 Moles of and 9 mol NaCl.

If 3 moles of NaCl reacts with one mole

Then 9 moles of NaCl require:

Hence, NaCl is considered as limiting reagent as it limits the formation of product.

So, 3 moles of Al (NO_3) _3 will react with 9 moles of NaCl and given moles of aluminium nitrate are 4, hence is present in excess and is considered as excess reagent.

9 mol NaCl to give 9 moles of.

Amount of 84.99 g/mol = 764.91 grams