A 14.570 g sample of CaCl2 was added to 12.285 g of K2CO3 and mixed in water. A 3.494 g yield of CaCO3 was obtained. What is the limiting reagent?
-CaCO3
-K2CO3
-CaCl2

Calculate the percent yield of CaCO3.
yield of CaCO3= %

Limiting reagent is the potassium carbonate.

Percent yield of calcium carbonate is: 39.3 %

Explanation:

The reaction is:

CaCl₂ + K₂CO₃ → CaCO₃ + 2KCl

Formula for percent yield is:

(Produced yield / Thoeretical yield) . 100

Firstly we determine the moles of each reactant, in order to say what is the limiting reagent: ratio is 1:1.

1 mol of chloride need 1 mol of carbonate.

14.570 g . 1 mol /110.98 g = 0.131 moles of CaCl₂

12.285 g . 1 mol / 138.2g = 0.0889 moles of carbonate.

Limiting reagent is carbonate. For 0.131 moles of CaCl₂ we need the same amount of carbonate and we have less moles.

Ratio is also 1:1, with calcium carbonate.

1 mol of potassium carbonate produces 1 mol of calcium carbonate

then, 0.0889 moles will produce the same amount of CaCO₃

We convert moles to mass: 0.0889 mol . 100.08g /mol = 8.89 g

That's the theoretical yield; to find the percent yield:

(3.494 g / 8.89g) . 100 = 39.3%

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i believe the answer is three water molecules