Bảo toàn khối lượng:

$m_O=13,1-5,19=7,91g$

$\Rightarrow n_O=\dfrac{7,91}{16}=0,494375(mol)$

$2H+O\to H_2O$

$\Rightarrow n_{H}=2n_O=0,494375.2=0,98875(mol)$

$\Rightarrow n_{H_2SO_4}=0,5n_H=0,494375(mol)$

$m_{dd H_2SO_4}=0,494375.98:24,5\%=197,75g$

$\to V_{dd}=\dfrac{197,75}{1,25}=158,2(ml)$

`m_(O)=13,1-5,19=7,91(g)`

`=>n_(O)=\frac{7,91}{16}=0,4943755(mol)`

$4A+nO_2\xrightarrow{t^o}2A_2O_n$

$4B+mO_2\xrightarrow{t^o}2B_2O_m$

`A_2O_n+nH_2SO_4->A_2(SO_4)_n+nH_2O`

`B_2O_m+mH_2SO_4->B_2(SO_4)_m+mH_2O`

Theo `PT`

`n_(H_2SO_4)=n_(H_2O)=n_(O)=0,494375(mol)`

`=>m_(H_2SO_4)=0,494375.98=48,44975(g)`

`m_(dd)=\frac{48,44975}{24,5%}193,795(g)`

`v_(dd)=\frac{193.795}{1,25}=155,036(ml)`