nFeS = \(\dfrac{m}{M}=\dfrac{8,8}{\left(56+32\right)}\) = 0.1 (mol)
nFeS2 = \(\dfrac{m}{M}=\dfrac{12}{\left(56+32.2\right)}\) = 0.1 (mol)
=> nS = 0.3 = nSO2
PTHH: SO2 + 2NaOH → Na2SO3 + H2O
TPT: 1mol 2mol
TĐB: 0.3mol → ?(mol)
=> nNaOH = \(\dfrac{0,3.2}{1}\) = 0,6(mol)
=> mNaOH = n . M = 0.6 . 40 = 24 (g)
Ta có: C% = \(\dfrac{m_{ctan}}{m_{\text{dd}}}\).100 => mdd = \(\dfrac{m_{ctan}}{C\%}.100\%\)
=> mdd = \(\dfrac{24}{25\%}.100\%\) = 96 (g)
Ta có: m = D . V
=> V = \(\dfrac{m}{D}\) = \(\dfrac{96}{1,28}\) = 75 ml