Calcium carbonate (100.1 g/mol) and hydrochloric acid (36.5 g/mol) react to form

Question

Calcium carbonate (100.1 g/mol) and hydrochloric acid (36.5 g/mol) react to form calcium chloride (111.0 g/mol), water and carbon dioxide in the reaction shown below.

CaCO3(s) + 2 HCl(aq) à CaCl2(aq) + H2O(l) + CO2(g)

a) What is the theoretical yield of calcium chloride when 25.0 g of calcium carbonate are combined with 13.0 g of hydrochloric acid?

b) Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

c) What would be the percent yield if 16.7 g of calcium chloride were isolated at the end of the reaction?

Explanation / Answer

CaCO3(s) + 2 HCl(aq) ---------> CaCl2(aq) + H2O(l) + CO2(g)

Molar mass(g/mol) 100.1 36.5 111

From the balanced reaction,

100.1 g of CaCO3 reacts with 2x36.5 g of HCl

M g of CaCO3 reacts with 13.0 g of HCl

M = ( 13.0x100.1) / (2x36.5)

= 17.8 g

So 25.0 - 17.8 = 7. g of CaCO3 left unreacted so which is the excess reactant.

Since all the mass of HCl completly reacted it is the limiting reactant.

Again from the equation,

2x36.5 g of HCl produces 111 g of CaCl2

13.0 g of HCl produces N g of CaCl2

N = ( 13.0x111)/(2x36.5)

= 19.8 g of CaCl2 ---> This is the theoretical yield

Percent yield = ( actual yield / theoretical yield) x100

= (16.7g / 19.8g) x 100

= 84.5 %

Therefore the percent yield is 84.5 %

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