Question

1.84 g of caco3 and mgco3 were treated with 50 ml of 0.8M HCl. Calculate the mass of CaCO3 mixture​

Answer 1

Answer:

Explanation:

Ew(CaCO3)=1002=50(n=2)  

Ew(MgCO3)=842=42(n=2)

Let x g of CaCO3 and (1.84−x)g of MgCO3 be present.

mEq. Of CaCO3+m Eq of MgCO3=m Eq of HCl

(x50+1.84−x42)×103=50×0.8×1

∴x=1g

%ofCaCO3=1×1001.84=54.35%

%ofMgCO3=100−54.35=45.65%