\(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:4M + 4nH2SO4 --> 2M2(SO4)n + 2nSO2 + 4nH2O
___\(\dfrac{0,6}{n}\)<-----------------------------------0,3
=> \(M_M=\dfrac{19,2}{\dfrac{0,6}{n}}=32n\left(g/mol\right)\)
Xét n = 1 => L
Xét n = 2 => MM = 64 (Cu)
=> C