$Cl_{2} + 2NaOH\rightarrow NaCl +NaClO +H_{2}O$ 0,05 mol 0,1mol$n_{Cl_{2}}=\frac{1,12}{22,4}=0,05 (mol)$$C_{M}=\frac{n}{V} \Rightarrow V_{NaOH}=\frac{n}{C_{M}}=\frac{0,1}{1} =0,1 lít $
Câu 1: $Cl_{2} + 2NaOH\rightarrow NaCl +NaClO +H_{2}O$ 0,05 mol 0,1mol$n_{Cl_{2}}=\frac{1,12}{22,4}=0,05 (mol)$$C_{M}=\frac{n}{V} \Rightarrow V_{NaOH}=\frac{n}{C_{M}}=\frac{0,1}{1} =0,1 lít $