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26 September, 10:35

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

Reaction Data

Reactants Products

Al (NO3) 3 NaCl NaNO3 AlCl3

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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  1. 26 September, 11:51
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    First thing you need to do is write the chemical equation and check to see if it is balancedAl (NO₃) ₃ + NaCl ⇒ NaNO₃ + AlCl₃The left hand side requires 3 more chloride ions, while the right hand side needs 3 nitrate ions to be balanced. Add a coefficient of 3 on NaCl and NaNO₃ to get a balanced equation. Al (NO₃) ₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃The ratio of Al (NO₃) ₃: NaCl is 1:3, which means 1 mol of Al (NO₃) ₃ reacts with 3 moles of NaCl ˣ = Clearly, NaCl is the limiting reactant since according to the ratio 1:3, 4 mol Al (NO₃) ₃ should react with 12 mol NaCl, but the reaction data has 9 mol of NaCl available. Use the limiting reactant moles to find the product NaNO₃ formed. The molecular weight of NaNO₃ is 84.9965 g/mol and thus mass of NaNO₃ formed is mass = (9 moles NaNO₃) (84.9965 g/mol NaNO₃) = 764.97g
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