Ta có:
\(n_{Cl2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{NaOH}=100.16\%=16\left(g\right)\)
\(\Rightarrow n_{NaOH}=\frac{16}{40}=0,4\left(mol\right)\)
Phản ứng xảy ra:
\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)
Vì nNaOH > 2nCl2 nên NaOH dư.
\(n_{NaCl}=n_{Cl2}=0,15\left(mol\right)\)
BTKL,
\(m_{ddX}=m_{Cl2}+m_{dd\left(NaOH\right)}=0,15.71+100\)
\(=110,65\left(g\right)\)
\(\Rightarrow C\%_{NaCl}=\frac{0,15.58,5}{110,65}.100\%=7,93\%\)
\(n_{Cl_2}=0,15\left(mol\right);n_{NaOH}=0,4\left(mol\right)\)
\(2NaOH+Cl_2\rightarrow NaCl+NaClO\)
0,3_______0,15____0,15
\(\Rightarrow m_{NaCl}=8,775\left(g\right);m_{dd\text{ }X}=110,65\left(g\right)\\ \Rightarrow C\%\left(NaCl\right)=7,93\%\)
