1. To calculate the limiting reagent, we need to calculate the number of moles of each reagent using their molar masses:
- K2CO3: 34.5 g / 138.21 g/mol = 0.25 mol
- HCl: 22.5 g / 36.46 g/mol = 0.62 mol
The stoichiometric ratio of K2CO3 to HCl is 1:2, which means that 1 mole of K2CO3 reacts with 2 moles of HCl. Therefore, the K2CO3 is the limiting reagent because 0.25 mol of K2CO3 requires 0.5 mol of HCl to react completely, but we only have 0.62 mol of HCl available.
2. To calculate the theoretical yield of H2O, we need to use the stoichiometric ratio of the balanced equation to find the number of moles of H2O that should be produced:
- 1 mol of K2CO3 reacts with 1/2 mol of H2O
- 0.25 mol of K2CO3 should react with 0.125 mol of H2O
- The molar mass of H2O is 18.02 g/mol, so the theoretical yield of H2O is:
0.125 mol x 18.02 g/mol = 2.25 g
3. To calculate the percent yield, we divide the actual yield (3.4 g) by the theoretical yield (2.25 g) and multiply by 100:
Percent yield = (actual yield / theoretical yield) x 100%
Percent yield = (3.4 g / 2.25 g) x 100%
Percent yield = 151.11%
The percent yield is greater than 100% which is not possible in actual lab situations. It indicates that there may have been errors in the experiment, such as incomplete reaction or loss of product during the experiment.