nSO2 =\(\dfrac{7,84}{64}=0,1225\left(mol\right)\)
nNaOH= 0,5 . 0,35=0,175(mol)
ta có tỉ lệ
\(\dfrac{nNaOH}{nSO2}=\dfrac{0,175}{0,1225}\simeq1,4\)
=> tạo hỗn hợp 2 muối
SO2 + NaOH -> NaHSO3
x ---------x
SO2 + 2NaOH -> Na2SO3 + H2O
y -------2y
ta có hpt :
\(\left\{{}\begin{matrix}x+y=0,1225\\x+2y=0,175\end{matrix}\right.\)
=> x= 0,07 , y = 0,0525
=> mNaHSO3 = 0,07 . 104=7,28(g)
mNa2SO3 =0,0525 . 126=6,615(g)
mNaHSO3+mNa2SO3 = 13,895(g)
b)_ CM NaHSO3 = \(\dfrac{0,07}{0,35}=0,2M\)
CM Na2SO3 =\(\dfrac{0,0525}{0,35}=0,15M\)