If a total of 13.5 mol of NaHCO3 and 4.5 mol of C6H8O7 react, how many moles of CO2 and Na3C6H5O7 will be produced?

3NaHCO3(aq) + C6H8O7(aq) > 3CO2(g) + 3H2O(s) + Na3C6H5O7(aq)

To find out how many moles of CO2 and Na3C6H5O7 will be produced, we need to use the coefficients from the balanced chemical equation.

From the balanced equation:
3 moles of NaHCO3 produce 3 moles of CO2 and 1 mole of Na3C6H5O7

Given:
13.5 mol of NaHCO3
4.5 mol of C6H8O7

Since the stoichiometric ratio between NaHCO3 and CO2 is 3:3, for every 3 moles of NaHCO3, 3 moles of CO2 are produced.

So, 13.5 mol of NaHCO3 will produce (13.5 / 3) * 3 = 13.5 moles of CO2

Similarly, since the stoichiometric ratio between C6H8O7 and Na3C6H5O7 is 1:1, 4.5 moles of C6H8O7 will produce 4.5 moles of Na3C6H5O7.

Therefore, when 13.5 mol of NaHCO3 and 4.5 mol of C6H8O7 react, it will produce:
13.5 moles of CO2
4.5 moles of Na3C6H5O7

To determine the number of moles of CO2 and Na3C6H5O7 produced, we first need to determine the limiting reactant.

The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reactant, we compare the moles of each reactant to their respective stoichiometric coefficients in the balanced equation.

Given:
- Moles of NaHCO3 = 13.5 mol
- Moles of C6H8O7 = 4.5 mol

Using the stoichiometric coefficients in the balanced equation:
- From the balanced equation: 3 mol NaHCO3 react with 1 mol C6H8O7 to produce 3 mol CO2 and 1 mol Na3C6H5O7.

Let's calculate the moles of CO2 that can be produced using each reactant:

For NaHCO3:
- Moles of CO2 = (13.5 mol NaHCO3) * (3 mol CO2 / 3 mol NaHCO3) = 13.5 mol CO2

For C6H8O7:
- Moles of CO2 = (4.5 mol C6H8O7) * (3 mol CO2 / 1 mol C6H8O7) = 13.5 mol CO2

As we can see, both reactants can produce the same amount of moles of CO2.

Now, let's calculate the moles of Na3C6H5O7 produced:

For NaHCO3:
- Moles of Na3C6H5O7 = (13.5 mol NaHCO3) * (1 mol Na3C6H5O7 / 3 mol NaHCO3) = 4.5 mol Na3C6H5O7

For C6H8O7:
- Moles of Na3C6H5O7 = (4.5 mol C6H8O7) * (1 mol Na3C6H5O7 / 1 mol C6H8O7) = 4.5 mol Na3C6H5O7

Again, both reactants can produce the same amount of moles of Na3C6H5O7.

Therefore, in this reaction, 13.5 moles of CO2 and 4.5 moles of Na3C6H5O7 will be produced.

This is a limiting reagent problem.

Step 1. Convert reagent 1 to mols of product.
Step 2. Convert reagent 2 to mols of product.
3. The smaller value (if there is a smaller value) will the number of mols CO2 produced.
Do the same for the salt.