Question

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? 3NaCO3(aq)+C6H8O7- ->3CO2(g)+3H2O(s)+Na3C6H5O7(aq)

Answer 1

The given chemical reaction is

3NaCO3(aq) + C6H8O7 ------------> 3CO2(g) + 3H2O(s) + Na3C6H5O7(aq)

For this we need to find the limiting reagent.

In order to find the limiting reagent, we need to divide the given moles with the respective stoichiometric coefficients. Then, the one with lowest number of moles is the limiting reagent.

Moles of NaHCO3 = 13.5 mol

Stoichiometric coefficient = 3

Moles = 13.5 / 3 = 4.5 mol

Moles of C6H8O7 = 4.5 mol

Stoichiometric coefficient = 1

Moles = 4.5 / 1 = 4.5 mol

So, both of them are equal. So, there is no limiting reagent, both are present in equal amounts. So, we can use any of them.

Thus, from the chemical reaction, it is clear that 1 mol of C6H8O7 gives 3 moles of CO2.

Thus, 4.5 mol of C6H8O7 gives (4.5 x 3) = 13.5 mol of CO2

And from chemical reaction, it is clear that 1 mol of C6H8O7 gives 1 mol of NaC6H5O7

Thus, 4.5 mol of C6H8O7 gives 4.5 mol of NaC6H5O7