Gọi số \(\left(g\right)\text{ }dd\text{ }NaOH\) cần dùng là \(x\left(g\right)\left(0< x< 224\right)\)
Số \(\left(g\right)\text{ }dd\text{ }Ba\left(OH\right)_2\) cần dùng là \(y\left(g\right)\left(0< y< 224\right)\)
\(m_{NaOH}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{x\cdot20}{100}=0,2x\left(g\right)\)
\(m_{Ba\left(OH\right)_2}=\dfrac{m_{d^2}\cdot C\%}{100}=\dfrac{y\cdot8,55}{100}=0,0855y\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{m}{M}=\dfrac{0,2x}{40}=0,005x\left(mol\right)\\ n_{Ba\left(OH\right)_2}=\dfrac{m}{M}=\dfrac{0,0855y}{171}=0,0005y\left(mol\right)\)
\(V_{d^2\text{ }HNO_3}=\dfrac{m}{D}=\dfrac{224}{1,12}=200\left(ml\right)\\ \Rightarrow n_{HNO_3}=n\cdot V=4,5\cdot0,2=0,9\left(mol\right)\)
\(pthh:NaOH+HNO_3\rightarrow NaNO_3+H_2O\left(1\right)\\ 0,005x.........0,005x\\ Ba\left(OH\right)_2+2HNO_3\rightarrow Ba\left(NO_3\right)_2+2H_2O\left(2\right)\\ 0,0005y.....0,001y\)
Ta có hệ pt:
\(\left\{{}\begin{matrix}0,005x+0,001y=0,9\\x+y=224\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=169\\y=55\end{matrix}\right.\left(T/m\right)\)
\(\Rightarrow m_{dd\text{ }NaOH}=169\left(g\right)\\ m_{dd\text{ }Ba\left(OH\right)_2}=55\left(g\right)\)
