Question

The following data was collected when a reaction was performed experimentally in the laboratory.
Determine the maximum amount of AlCl3 that was produced during the experiment. Explain how you determined this amount.

Answer 1

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Answer 2

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Answer 3

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Answer 4

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Answer 5

reactants are Al(NO3)3 + NaCl
products are NaNO3 + AlCl3

balanced equation: Al(NO3)3 + 3NaCl --> AlCl3 + 3NaNO3
(please review balancing equations if you are unsure how this is derived)

now, you're given 4 moles Al(NO3)3 and 9 moles NaCl, so you'll need to determine which one is the limiting reactant.

use the mole ratio in the chemical equation to convert:

1. 4 moles Al(NO3)3 to moles AlCl3
2. 9 moles NaCl to moles AlCl3

whichever gives you the lowest amount of AlCl3 is the limiting reactant. the lowest # of moles AlCl3 is the maximum amount of AlCl3 and this goes in the product section.

now, looking at the ratio in the equation: 1 mol AlCl3 = 3 moles NaNO3. so take the moles AlCl3 you determined in the previous step, and multiply by 3 to get moles NaNO3.

Answer 6

\[Al(NO_{3})_{3}+ 3NaCl \rightarrow 3 Na NO_{3} + AlCl_{3} ~~~...(1)\]
so 1 mole of \[Al(NO_{3})_{3}\] with 3 moles of NaCl.
so 9 moles of NaCl combine with 3 moles Of \[Al(NO_{3})_{3}\]
multiply (1) by 3
\[3 Al(NO_{3})_{3}+9 NaCl \rightarrow 9NaNO_{3}+3AlCl_{3}\]