\(n_{NaCl}=0,4\left(mol\right)\)
\(n_{NaOH}=0,3\left(mol\right)\)
TH1 : NaOH hết
\(\Rightarrow n_{NaCl}=n_{NaClO}=0,15\left(mol\right)\)
\(\Rightarrow m_r=m_{NaCl}+m_{NaClO}=43,35\left(g\right)\)
\(\Rightarrow\) Dư NaOH
\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)
2x______x___________x____x_____________
\(\Rightarrow Spu:0,3-2x\left(mol\right)NaOH\)
\(0,4+x\left(mol\right)NaCl\)
\(x\left(mol\right)NaClO\)
\(\Rightarrow40\left(0,3-2x\right)+58,5\left(0,4+x\right)+74,5=49,85\)
Bạn giải pt tìm x. Tính V= 22,4x