\(n_{MnO_2}=\dfrac{69,6}{87}=0,8\left(mol\right)\\ MnO_2+4HCl_{đặc,nóng}\rightarrow MnCl_2+Cl_2+2H_2O\\ Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\\ n_{Cl_2}=n_{MnO_2}=0,8\left(mol\right)\\ n_{NaCl}=n_{NaClO}=n_{Cl_2}=0,8\left(mol\right)\\ V_{dd.muoi}=V_{ddNaOH}=0,5\left(l\right)\\ C_{MddNaCl}=\dfrac{0,8}{0,5}=1,6\left(M\right)\\ C_{MddNaClO}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)
Đúng 2
Bình luận (1)
