Question

Answer 1

PTHH:
CaCO3+2HCl--->CaCl2+H2O+CO2(1)
x__________________________x
MgCO3+2HCl---->MgCl2+H2O+CO2(2)
y_________________________y
CO2+Ba(OH)2--->BaCO3+H2O(3)
0,03__0,03________0,03
2CO2+Ba(OH)2---->Ba(HCO3)2 (4)
0,03____0,015
n_{Ba(OH)2 bđ}=0,045 mol
n_BaCO3=0,03 mol
Theo pt (3): n_CO2=n_BaCO3=0,03 mol
=> n_{Ba(OH)2 pu}=0,03 mol
=>n_{Ba(OH)2 dư}=0,015 mol
=>n_CO2(4)=0,03 mol
=>\(\Sigma\) n_CO2=0,06 mol
Đặt n_CaCO3=x mol;n_MgCO3=y mol
Ta có hệ pt
\(\left\{{}\begin{matrix}100x+84y=5,68\\x+y=0,06\end{matrix}\right.\)
=> x=0,04 ; y=0,02

=> %CaCO3= 70,42% ; %MgCO3= 29,58%