Gọi \(CT:R\left(HCO_3\right)n\)
\(2R\left(HCO_3\right)n+nH_2SO_4\rightarrow R_2\left(SO_4O_4\right)n+2nCO_2+2nH_2O\)
\(9,875g...............8,25gam\)
Ta có : \(\dfrac{9,875}{\left(R+61n\right)}=\dfrac{8,25.2}{\left(2R+96n\right)}\)
\(\rightarrow3,25R=58,5n\rightarrow R=18n\)
\(n=1\rightarrow R=18\rightarrow\) R là \(NH_4\)
Vậy CT muối : \(NH_4HCO_3\)