\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
Ta có :
\(n_{MnO2}=\frac{10,44}{87}=0,12\left(mol\right)\)
\(n_{Cl2}=n_{MnO2}=0,12\left(mol\right)\)
\(n_{NaOH}=\frac{60.20\%}{40}=0,3\left(mol\right)\)
\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
\(n_{NaOH.trong.Y}=0,3-0,12.2=0,06\left(mol\right)\)
\(n_{NaCl}=n_{NaClO}=0,12\left(mol\right)\)
\(m_{dd\left(Spu\right)}=0,12.71+60=68,52\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaOH}=3,5\%\\C\%_{NaCl}=10,25\%\\C\%_{NaClO}=13,05\%\end{matrix}\right.\)
