CHÚC BẠN HỌC TỐT
1) Theo đề bài, ta có: \(\left\{{}\begin{matrix}m_{NaOH}=200.1,15.8\%=18,4\left(g\right)\\m_{MgCl2}=380.5\%=19\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=\dfrac{18,4}{40}=0,46\left(mol\right)\\n_{MgCl2}=\dfrac{19}{95}=0,2\left(mol\right)\end{matrix}\right.\)
PTHH: \(2NaOH+MgCl_2\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)
pư.............0,4..............0,2...............0,4............0,2 (mol)
Ta có tỉ lệ: \(\dfrac{0,46}{2}>\dfrac{0,2}{1}\) Vậy NaOH dư, MgCl2 hết.
\(\Rightarrow m_{NaOHdư}=40.\left(0,46-0,4\right)=2,4\left(g\right)\)
2) \(\Rightarrow m_{Mg\left(OH\right)2}=0,2.58=11,6\left(g\right)\)
\(m_{dds}=m_{ddNaOH}+m_{ddHCl}-m_{Mg\left(OH\right)2}\)
\(\Rightarrow m_{dds}=200.1,15+380-11,6=598,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,4.58,5}{598,4}.100\%\approx3,91\%\\C\%_{NaOHdư}=\dfrac{2,4}{598,4}.100\%\approx0,4\%\end{matrix}\right.\)
Vậy............
