For the first problem: First calculate the molar mass of KClO_{3}:

K: 39.098

Cl: 35.446

O: 3x15.999 = 47.997

KClO_{3}: 122.541 g/mol

So 23.5 g = 23.5/122.541 = 0.19177 moles of KClO_{3}

Using the molar ratios, there are 2 moles of KClO_{3} required for every 2 moles of 2KCl produced meaning that there are 0.19177 moles of KCl produced.

Then calculate the molar mass of KCl:

K: 39.098

Cl: 35.446

KCl: 74.544 g/mol

Therefore 0.19177 moles of KCl times 74.544 g/mol = 14.3 grams.of HCl

Since 9.32 g of KCl, the percent yield was 9.32/14.3 = 65.2%

The other problems are similar. For the problems regarding the limiting reactant, see which of the reactants has excess (more than what is needed) and then the other reactant will be the limiting reactant and can be used to calculate the results.