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28 June, 10:57

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.

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Answers (1)
  1. 28 June, 11:30
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    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
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