Question

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

Answer 1

Given - Number of moles of Al(NO3)3 - 4 moles

Number of moles of NaCl - 9 moles

Find - Maximum amount of AlCl3 produced during the reaction.

Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3

To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.

Mole ratio Al(NO3)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 AlCl3

9 moles of NaCl will produce - 1/3*9 - 3 moles.

Weight of AlCl3 - 3*133.34 - 400 grams

Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Answer 2

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