\(n_{H_2SO_4}=\dfrac{140\cdot9.8\%}{98}=0.14\left(mol\right)\)
\(M_2\left(CO_3\right)_n+nH_2SO_4\rightarrow M_2\left(SO_4\right)_n+nCO_2+nH_2O\)
\(................0.14.......\dfrac{0.14}{n}\)
\(M_{Muối}=\dfrac{19.04}{\dfrac{0.14}{n}}=136n\)
\(\Rightarrow2M+96n=136n\)
\(\Rightarrow M=20n\)
\(BL:n=2\Rightarrow M=40\)
\(CT:CaCO_3\)