Đáp án:
$C\%_{Na_2SO_3}=1\%$
$C\%_{NaHSO_3}=1,23\%$
Giải thích các bước giải:
$n_{SO_2}=\dfrac{0,896}{22,4}= 0,04 mol$
$n_{NaOH}=\dfrac{200.1,12\%}{40}=0,056 mol$
$\dfrac{n_{NaOH}}{n_{SO_2}}=1,4\to$ tạo $Na_2SO_3$ (x mol), $NaHSO_3$ (y mol)
$2NaOH+SO_2\to Na_2SO_3+H_2O$
$NaOH+SO_2\to NaHSO_3$
$\Rightarrow 2x+y=0,056; x+y=0,04$
$\Rightarrow x=0,016; y=0,024$
$m_{dd}=0,04.64+200=202,56g$
$C\%_{Na_2SO_3}=\dfrac{0,016.126.100}{202,56}=1\%$
$C\%_{NaHSO_3}=\dfrac{0,024.104.100}{202,56}=1,23\%$