$a) Cl2 + 2NaOH \rightarrow NaCl+ NaClO+ H2O$ $a 2a a mol$
$giả sử NaOH p/ư hết:\rightarrow a=0,09 mol\rightarrow m chất tan=m NaCl+m NaClO=0,09(23+35,5+23+35,5+16)=11,97g khác 10,38g\rightarrow NaOH dư$
$có 2a là số mol NaOH p/ư\rightarrow m chất tan=m NaCl+ m NaClO+ m NaOH dư=40(0,18-2a)+a(35,5+23+35,5+23+16)=10,38\rightarrow a=0,06 mol\rightarrow V=0,06.22,4=1,344l$
$b)2NaCl+ 2H2O\overset{đp có màng ngăn}{\rightarrow}2NaOH + Cl2 + H2$
$0,12 0,06$
$\rightarrow H=80\rightarrow n NaCl cần lấy=0,15 mol\rightarrow m NaCl=8,775g\rightarrow m dd NaCl=8,775.100/30=29,25g$