$C_M(KOH)=\dfrac{10dC\%}{M}=\dfrac{10 \times 1.41 \times 40}{56}(M)$

`->`$n_{KOH}=\dfrac{10 \times 1.41 \times 40}{56} \times \dfrac{74.5}{1000}=0.75(mol)$

$H_2SO_4+2KOH \to K_2SO_4+2H_2O$

`=>`$n_{H_2SO_4}=\dfrac{n_{KOH}}{2}=0.375(mol)$$(*)$

$n_{H_2SO_4.kSO_3}=\dfrac{32.25}{98+80k}(mol)$

$H_2SO_4.kSO_3+(k+1)H_2O \to (k+1)H_2SO_4$

`=>`$n_{H_2SO_4}=(k+1) \times \dfrac{32.25}{98+80k}(mol)$$(**)$

Từ $(*),(**)$ suy ra: $(k+1) \times \dfrac{32.25}{98+80k}=0.375$

`=>`$k=2$

Vậy công thức của oleum cần tìm là $H_2SO_4.2SO_3$

`X:H_2SO_4.nSO_3`

`n_{KOH}={74,5.1,41.40\%}/{56}=0,75(mol)`

`H_2SO_4.nSO_3+nH_2O->(n+1)H_2SO_4`

`H_2SO_4+2KOH->K_2SO_4+2H_2O`

`->n_X={n_{KOH}}/{2(n+1)}={0,375}/{n+1}(mol)`

`->M_X=98+80n={32,25}/{{0,375}/{n+1}}=86(n+1)`

`->n=2`

Vậy oleum là `H_2SO_{4}.2SO_3`