Question

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?

Answer 1

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.

Answer 2

Explanation:

1. . In the reaction,
2. CaCO 2HCl CaCl CO H O 3 2 2 2    
3. 6.088 g
4. CaCO3
5. reacted with 2.852 g HCl. What mass of
6. CaCO3
7. remains unreacted