a, \(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
b,
\(n_{MnO2}=0,4\left(mol\right)\Rightarrow n_{Cl2}=0,4\left(mol\right)\)
\(n_{NaOH}=1\left(mol\right)\)
Sau phản ứng thu được : \(n_{NaCl}=n_{NaClO}=0,4\left(mol\right)\)
\(n_{NaOH_{dư}}=0,2\left(mol\right)\)
\(\Rightarrow CM_{NaCl}=CM_{NaClO}=0,8M\)
\(\Rightarrow CM_{NaOH}=0,4M\)
a)
MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O (1)
Cl2 + 2NaOH -> NaCl + NaClO + H2O (2)
b) n MnO2= 34.8/87=0.4 (mol)
n NaOH= 0.5*2=1 (mol)
(1) => n Cl2= n MnO2= 0.4 (mol)
Theo phương trình (2), ta thấy:
\(\frac{0.4}{1}< \frac{1}{2}\) => NaOH dư
(2) suy ra:
+ n NaCl = n NaClO = n Cl2 = 0.4 (mol)
+ n NaOH phản ứng = 2n Cl2 = 0.8 (mol ) => n NaOH dư= 1-0.8=0.2 (mol)
Dung dịch sau phản ứng gồm: NaCl, NaClO, NaOH dư
\(C_{M_{NaCl}}\) =0.4/0.5=0.8M
\(C_{M_{NaClO}}\)=0.4/0.5=0.8M
\(C_{M_{NaOHdư}}\)=0.2/0.5=0.4M