MgCO3+H2SO4\(\rightarrow\)Mg2SO4 + H2O+CO2
MgO+H2SO4\(\rightarrow\)Mg2SO4 + H2O
b) n(CO2)=\(\frac{3,36}{22,4}\)=0,15 mol
m(MgCO3)=0,15.(24+60) = 12,6g
%m(MgCO3)=\(\frac{12,6}{16,6}\).100=75,9%
m(MgO) =16,6-12,6=4g
%m(MgCO3) =\(\frac{4}{16,6}\) .100 =24,1%
c) n(MgO)=\(\frac{4}{24+16}\)=0,1 mol
m muối = m (Mg2SO4)=(0,1+0,15). (24.2 + 96)= 36g