Question

NEED HELP ASAP NOT DIFFICULT

Answer 1

Answer:

9 moles of NaNO3 and 3 moles of AlCl3

Explanation:

First, we need the balanced equation, which is:

Al(NO3)3 + (3)NaCl ---> (3)NaNO3 + AlCl3

Next, we need the limiting reactant, which is NaCl because we require more of it than Al(NO3) to make AICl

Now we need to determine the maximum of AlCl we can produce:

Because we need 3 moles of NaCl to react with Al(NO3)3 to produce 1 mole of AlCl3, we can only produce 3 AlCl3 due to insufficient NaCl.

So we would have 9 NaNO3 and 3 AlCl3.

Answer 2

Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

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.