MgCO3+2HCl--->MgCl2+H20+CO2
x---------------------------------------1\2x mol
Na2CO3+2HCl-->2NaCl+H2O+CO2
y--------------------------------------y mol
nCO2=\(\frac{4,48}{22,4}\)=0,2 mol
ta có hệ pt 84x+106y=19
x+y =0,2
==>x=0,1 mol ;y=0,1 mol
tacó %mMgCO3=\(\frac{84.0,1}{19}.100\text{%}\)=44,2%
%mNa2CO3=100-44,2=55,8%
b)bạn tự tính nhé