`-` TN1: Chọn \(m_{ddA}=m_{ddB}=100\left(g\right)\)
`=>` \(\left\{{}\begin{matrix}n_{HNO_3}=\dfrac{100.a\%}{63}=\dfrac{a}{63}\left(mol\right)\\n_{Ba\left(OH\right)_2}=\dfrac{100.b\%}{171}=\dfrac{b}{171}\left(mol\right)\end{matrix}\right.\)
Ta có: \(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\)
PTHH: \(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)
\(\dfrac{a}{126}\)<------\(\dfrac{a}{63}\)
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
0,1<---------0,1
`=>` \(n_{Ba\left(OH\right)_2\left(dư\right)}=\dfrac{b}{171}-\dfrac{a}{126}=0,1\left(1\right)\)
`-` TN2: Chọn \(\left\{{}\begin{matrix}m_{ddA}=200\left(g\right)\\m_{ddB}=100\left(g\right)\end{matrix}\right.\)
`=>` \(\left\{{}\begin{matrix}n_{HNO_3}=\dfrac{200.a\%}{63}=\dfrac{2a}{63}\left(mol\right)\\n_{Ba\left(OH\right)_2}=\dfrac{100.b\%}{171}=\dfrac{b}{171}\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\)
\(\dfrac{b}{171}\)------>\(\dfrac{2b}{171}\)--------->\(\dfrac{b}{171}\)
`=>` \(\left\{{}\begin{matrix}m_{Ba\left(NO_3\right)_2}=\dfrac{261b}{171}=\dfrac{29b}{19}\left(g\right)\\m_{HNO_3\left(dư\right)}=63.\left(\dfrac{2a}{63}-\dfrac{2b}{171}\right)=2a-\dfrac{14b}{19}\left(g\right)\end{matrix}\right.\)
`=>` \(\dfrac{m_{Ba\left(NO_3\right)_2}}{m_{HNO_3}}=\dfrac{\dfrac{29b}{19}}{2a-\dfrac{14b}{19}}=\dfrac{87}{14}\)
`=>` \(\dfrac{406b}{19}=174a-\dfrac{1218b}{19}\)
`=>` \(174a-\dfrac{1624b}{19}=0\left(g\right)\)
Từ `(1), (2) =>` \(\left\{{}\begin{matrix}a=25,2\%\\b=51,3\%\end{matrix}\right.\)
