PTHH : \(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)
\(n_{Cl2}=\frac{V_{Cl2}}{22,4}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
- Theo PTHH : \(n_{NaOH}=2n_{Cl2}=2.0,05=0,1\left(mol\right)\)
=> \(V_{NaOH}=\frac{n_{NaOH}}{C_{MNaOH}}=\frac{0,1}{0,5}=0,2\left(l\right)\)
nCl2 = 0.05 mol
2NaOH + Cl2 => NaCl + NaClO + H2O
0.1______0.05
VddNaOH = 0.1/0.5 = 0.2 l