Question

Reaction Data ReactantsProductsAl(NO3)3NaClNaNO3AlCl3Starting Amount in Reaction4 moles9 moles??Determine the maximum amount of NaNO3 that was produced during the experiment. Explain how you determined this amount. I've had help on this before but im still struggling.

Answer 1

Step 1: Write and balance the equation:

3 NaCl + Al(NO3)3 → AlCl3 + 3 NaNO3

Step 2: We need to find the limiting and excess reactant. For this, we need to look at the stoichiometric ratio between the reactants:

3 moles of NaCl reacts with 1 mole of Al(NO3)3.

Now we do a rule of 3 for both.

For 9 moles of NaCl:

3 moles of NaCl --- 1 mole of Al(NO3)3

9 moles of NaCl --- x mole of Al(NO3)3

3x = 9

x = 3 moles of Al(NO3)3

For 4 moles of Al(NO3)3:

3 moles of NaCl --- 1 mole of Al(NO3)3

x moles of NaCl --- 4 moles of Al(NO3)3

x = 12 moles of NaCl

As we can see, to react with 4 moles of Al(NO3)3 we should have 12 moles of NaCl, but we have just 9. So the NaCl is the limiting reactant and the excess reactant is Al(NO3)3.

Step 3: We need to look at the stoichiometric ratio between the limiting reactant (NaCl) and NaNO3:

3 moles of NaCl --- 3 moles of NaNO3

9 moles of NaCl --- x moles of NaNO3

x = 9 moles of NaNO3

Answer: The maximum amount of NaNO3 that was produced during the experiment is 9 moles.