MgCO3---->MgO+CO2
x--------------------x--x
CaCO3------->CaO+CO2
y------------------y---------y
n CO2=1,68/22,4=0,075(mol)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}40x+56y=3,8\\x+y=0,075\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,025\\y=0,05\end{matrix}\right.\)
m hh=3,8+0,075.44=7,1(g)
%m MgCO3=0,025.84/7,1.100%=29,58%
%m CaCO3=100%-29,58=70,42%
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