3.50 mol Al(NO3)3 reacts with 5.00 mols K2CrO4, the total number of moles of Al2(CrO4)3 and KNO3 would be?

Here is the equation if it's needed.
2 Al(NO3)3(aq) + 3 K2CrO4(aq) → Al2(CrO4)3(s) + 6 KNO3(aq)

You do this the same way as the last one except this is a limiting reagent problem whereas the last one was not.

Use 3.50 mol Al(NO3)3 and determine mols KNO3.
Do it a second time with 5.00 mols K2CrO4 and calculate mols KNO3 formed.
It is likely the two answers will be different; the correct one in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Post your work if you still have questions and explain your trouble spots.

To find the total number of moles of Al2(CrO4)3 and KNO3 produced, we can use the stoichiometry of the balanced chemical equation.

From the balanced equation:
2 moles of Al(NO3)3 react with 3 moles of K2CrO4 to produce 1 mole of Al2(CrO4)3 and 6 moles of KNO3.

Given:
- 3.50 mol of Al(NO3)3
- 5.00 mol of K2CrO4

To find the limiting reactant, we compare the moles of each reactant based on their stoichiometric coefficients.

For Al(NO3)3:
3.50 mol Al(NO3)3 * (3 mol K2CrO4 / 2 mol Al(NO3)3) = 5.25 mol K2CrO4 required

For K2CrO4:
5.00 mol K2CrO4 * (2 mol Al(NO3)3 / 3 mol K2CrO4) = 3.33 mol Al(NO3)3 required

Since we have more K2CrO4 moles available (5.00 mol), Al(NO3)3 is the limiting reactant.

Now, let's determine the moles of products formed.

From the balanced equation:
1 mole of Al2(CrO4)3 is produced for every 2 moles of Al(NO3)3 reacted.

Therefore, the moles of Al2(CrO4)3 produced is:
3.50 mol Al(NO3)3 * (1 mol Al2(CrO4)3 / 2 mol Al(NO3)3) = 1.75 mol Al2(CrO4)3

From the balanced equation:
6 moles of KNO3 are produced for every 2 moles of Al(NO3)3 reacted.

Therefore, the moles of KNO3 produced is:
3.50 mol Al(NO3)3 * (6 mol KNO3 / 2 mol Al(NO3)3) = 10.50 mol KNO3

Therefore, the total number of moles of Al2(CrO4)3 and KNO3 formed is:
1.75 mol Al2(CrO4)3 + 10.50 mol KNO3 = 12.25 mol

To find the total number of moles of Al2(CrO4)3 and KNO3 produced, we need to use the stoichiometric coefficients from the balanced chemical equation you provided.

From the equation: 2 Al(NO3)3(aq) + 3 K2CrO4(aq) → Al2(CrO4)3(s) + 6 KNO3(aq)

We can see that the stoichiometric coefficient for Al2(CrO4)3 is 1, and the stoichiometric coefficient for KNO3 is 6.

Given that you have 3.50 moles of Al(NO3)3 and 5.00 moles of K2CrO4, we can determine the limiting reactant. This is the reactant that will be completely consumed and determine the maximum amount of product that can be formed.

To find the limiting reactant, we compare the ratios of moles of each reactant to their stoichiometric coefficients in the balanced equation.

For Al(NO3)3:
3.50 moles Al(NO3)3 × (1 mole Al2(CrO4)3 / 2 moles Al(NO3)3) = 1.75 moles Al2(CrO4)3

For K2CrO4:
5.00 moles K2CrO4 × (1 mole Al2(CrO4)3 / 3 moles K2CrO4) = 1.67 moles Al2(CrO4)3

Comparing the calculated moles of Al2(CrO4)3, we find that K2CrO4 is the limiting reactant because it produces fewer moles of Al2(CrO4)3.

Since we've established that K2CrO4 is the limiting reactant, we can determine the moles of KNO3 formed using its stoichiometric coefficient.

1.67 moles Al2(CrO4)3 × (6 moles KNO3 / 1 mole Al2(CrO4)3) = 10.02 moles KNO3

Therefore, the total number of moles of Al2(CrO4)3 and KNO3 produced would be 1.67 moles of Al2(CrO4)3 and 10.02 moles of KNO3.