In the reaction: CaCO3 + 2HCl CaCl2 + CO2 + H2O 6.088 gm of CaCO3 reacted with 2.852 gm of HCl. What is the limiting reagent and how much is CaCl2 produced?
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In the reaction: CaCO3 + 2HCl CaCl2 + CO2 + H2O 6.088 gm of CaCO3 reacted with 2.852 gm of HCl. What is the limiting reagent and how much is CaCl2 produced?
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Answer:
CaCO3 + 2HCl -----> CaCl2 + H2O + CO2.
2.5g? g
Mass of CaCO3= 40+12+3(16)=52+ 48= 100.
Molar mass of CaCO3=100g.
Mass of CaCl2= 40 + 2(35.5)= 40+ 71= 111.
Molar mass of CaCl2=111g.
Mass of CaCO3 (g) Mass of CaCl2 (g)
For 100g of CaCO
3
, 111g of CaCl_2$$ is formed.
let for 2.5g of CaCO
3
, x g of CaCl
2
is formed.
Thus, by cross multiplication,
x=111×2.5/100=2.775g=2.78g.
Explanation: