a,
$n_{SO_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$n_{KOH}=0,2.2=0,4(mol)$
$\to \dfrac{n_{KOH}}{n_{SO_2}}=1,33$
$\to$ tạo 2 muối: $KHSO_3$ ($x$ mol), $K_2SO_3$ ($y$ mol)
Bảo toàn $S$: $x+y=0,3$
Bảo toàn $K$: $x+2y=0,4$
$\to x=0,2; y=0,1$
Vậy $m_{\text{muối}}=0,2.120+0,1.158=39,8g$
b,
$m_{dd (1)}=1,5.1000.1,08=1620g$
$m_{dd(2)}=1.1000.1,12=1120g$
$\to m_{\text{dd sau khi trộn}}=1620+1120=2740g$
$m_{HNO_3\text{sau khi trộn}}=1620.12\%+1120.20\%=418,4g$
$\to C\%_{HNO_3}=\dfrac{418,4.100}{2740}=15,27\%$
c,
$S=36g$
$\to C\%_{\text{bão hoà}}=\dfrac{100S}{100+S}=26,47\%$
