Question

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

Answer 1

The maximum amount of $AlCl_3$ produced during the experiment can be find out by knowing the Limiting Reagent.

Here, The maximum amount of $AlCl_3$ produced during the experiment is  400 grams

What is Limiting reagent ?

The limiting reagent (or limiting reactant or limiting agent) in a chemical reaction is a reactant that is totally consumed when the chemical reaction is stopped.

Now, According to the question,

Given ;

Number of moles of $Al(NO_3)_3$ - 4 moles

Number of moles of NaCl - 9 moles

To Find - Maximum amount of $AlCl_3$ produced during the reaction.

Solution ;

complete reaction ;

$Al(NO_3)_3$  + 3NaCl -->   $3NaNO_3$ + $AlCl_3$

To find the maximum amount of $AlCl_3$ produced during the reaction,

we need to find the limiting reagent.

Mole ratio $Al(NO_3)_3$  = 4/1 = 4

Mole ratio NaCl = 9/3 = 3

Thus, NaCl is the limiting reagent in the reaction.

Now, 3 moles of NaCl produces 1 mole of   $Al(NO_3)_3$

9 moles of NaCl will produce = 1/3 x 9 = 3 moles.

Weight of   $Al(NO_3)_3$ = 3 x 133.34 = 400 grams

Thus, 400 grams of   $Al(NO_3)_3$  is the maximum amount of   $Al(NO_3)_3$ produced during the experiment.

Learn more about Limiting Reagent Here ;

brainly.com/question/20070272

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