How to determine theoretical yield

Question

KHCO3(aq)+HCl(aq) ———> KCl(aq)+H2O (l)+CO2(g)

Answer 1

The first step to these types of problems is to balance the entire equation.

Luckily this reaction's equation is balanced and each compound has a coefficient of 1 meaning that if you react one mole of KHCO3 with one mole of HCl you would produce one mole each of the reactants.

The next step is to determine the molar mass of each compound and you do this by adding the individual mass of each element in the compound to obtain a grand sum.

Molar Mass of KHCO3 = 100.115 g/mol

Molar Mass of HCl = 36.46 g/mol

Molar Mass of H2O = 18.01528 g/mol

Molar Mass of KCl = 74.5513 g/mol

Molar Mass of CO2 = 44.01 g/mol

Now that we know these values we can move on to the actual calculation. There is a problem; however, because you didn't specify if there is a limiting reagent or not. Limiting reagents are the compound that will be used up in the reaction first and stop the reaction from continuing. To determine the limiting reagent all you need to do is take the amount of each reactant and convert the amount to moles and the compound that has the lowest amount of moles is the limiting reagent.

You either start the calculation with the amount of the limiting reagent you have (see above) or you start out with whatever amount the questions asks for (i.e you have 0.8g of HCl and KHCO3 is in excess, blah, blah, how much water can be produced?).

I'm going to start the calculation with an arbitrary amount of KHCO3 and follow it all the way through for CO2 since the process is the exact same for determining the amounts of all the products.

(1.5 g of KHCO3) * (1 mol of KHCO3 / 100.115 g) * (1 mol CO2 / 1 mol KHCO3) * (44.01 g / 1 mol CO2 ) = 0.659 gram of CO2 produced.