What is the theoretical yield (in grams) of CaCO_3 in the precipitation reaction

Question

What is the theoretical yield (in grams) of CaCO_3 in the precipitation reaction that occurs when a solution prepared by dissolving 1.004 g of solid CaCl_2 in 30 mL of water is mixed with a solution prepared by dissolving 0.997 g of Na_2CO_3 in 30 mL of water? Suppose that 0.814 g of CaCO_3 is actually isolated in the precipitation reaction described above, What is the percent yield of this reaction? A solution of Na_2CO_3 is prepared by dissolving 0.511 g of the solid in approximately 30 mL water. How many moles of Na_2CO_3 does this solution contains? Suppose the Na_2CO_3 solution prepared by dissolving 0.511 g of the solid in approximately 30 mL of water is added to a solution that contains excess CaCl_2? What is the maximum number of moles of solid CaCO_3 that can form in the resulting precipitation reaction? What is the maximum mass (in grams) of CaCO_3 that could form in the precipitation reaction this time? True or false? CaCl_2 is the limiting reactant when a solution prepared by dissolving 1.004 g of solid CaCl_2 in 30 mL of water is mixed with a solution prepared by dissolving 0.511 g of Na_2CO_3 in 30 mL of water. What is the theoretical yield (in grams) of CaCO_3 in the precipitation reaction that occurs

Explanation / Answer

The balanced reaction is CaCl2 + Na2CO3 -------> CaCO3 + 2NaCl

Molar mass ( g/mol) 111 106 100

1.004 0.997 ?

From the balanced equation,

111 of CaCl2 reacts with 106 g of Na2CO3

1.004 g of CaCl2 reacts with M g of Na2CO3

M = ( 1.004 x 106) / 111

= 0.959 g

So 0.997 - 0.959 = 0.038 g of Na2CO3 left unreacted so Na2CO3 is the excess reactant

Since all the mass of CaCl2 completly reacted it is the limiting reactant

Again from the reaction,

111 g of CaCL2 produces 100 g of CaCO3

1.004 g of CaCl2 produces N g of CaCO3

N = ( 1.004 x 100) / 111

= 0.904 g of CaCO3 ---> This was the theoretical yield

Percent yield = ( actual yield / theoretical yield) x 100

= ( 0.814 / 0.904) x100

= 90.0 %

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