The percent yield of [tex]CaCO_{3}[/tex] is 36.6%.
To determine the limiting reagent, we need to compare the moles of each reactant to the stoichiometric ratio of the balanced chemical equation.
The balanced chemical equation for the reaction between CaCl2 and K2CO3 is:
[tex]CaCl_{2} + K_{2} CO_{3} → CaCO_{3} + 2KCl[/tex]
First, we calculate the number of moles for each reactant:
Moles of CaCl2 = mass / molar mass = 14.579 g / (40.08 g/mol + 2 * 35.45 g/mol) = 0.0971 mol
Moles of K2CO3 = mass / molar mass = 28.016 g / (2 * 39.10 g/mol + 12.01 g/mol + 3 * 16.00 g/mol) = 0.1370 mol
Next, we determine the stoichiometric ratio between CaCl2 and [tex]CaCO_{3}[/tex]:
According to the balanced equation, the ratio is 1:1. This means that 1 mole of CaCl2 reacts with 1 mole of [tex]CaCO_{3}[/tex].
Comparing the moles of CaCl2 (0.0971 mol) and K2CO3 (0.1370 mol), we can see that CaCl2 is the limiting reagent because there are fewer moles of CaCl2 compared to the stoichiometric ratio.
To calculate the percent yield of [tex]CaCO_{3}[/tex], we use the actual yield and theoretical yield. Theoretical yield is the maximum amount of CaCO3 that could be produced based on the limiting reagent.
The molar mass of [tex]CaCO_{3}[/tex]is 100.09 g/mol. Since the balanced equation shows a 1:1 ratio between CaCl2 and CaCO3, the moles of [tex]CaCO_{3}[/tex]formed will be the same as the moles of CaCl2.
The theoretical yield of [tex]CaCO_{3}[/tex] is: Moles of CaCO3 = Moles of CaCl2 = 0.0971 mol
Mass of [tex]CaCO_{3}[/tex] (theoretical) = Moles of [tex]CaCO_{3}[/tex] * Molar mass of CaCO3 = 0.0971 mol * 100.09 g/mol = 9.72 g
The percent yield of CaCO3 is calculated as:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (3.558 g / 9.72 g) * 100 = 36.6%
Therefore, the percent yield of CaCO3 is 36.6%.
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