Đáp án:

Giải thích các bước giải:

$n_{NaBr}=0,1.0,8=0,08(mol)\\ n_{NaI}=0,1.0,06=0,006(mol)\\ a,\\ Cl_2+2NaBr\to 2NaCl+Br_2\\ Cl_2+2NaI\to 2NaCl+I_2\\ \xrightarrow{\text{Từ PTHH}} \ \begin{cases}n_{Cl_2}=\dfrac{1}{2}.(n_{NaBr}+n_{NaI})=0,043(mol\\n_{NaCl}=2.n_{Cl_2}=0,086(mol)\\ n_{Br_2}=\dfrac{1}{2}.n_{NaBr}=0,04(mol)\\n_{I_2}=\dfrac{1}{2}.n_{NaI}=0,003(mol)\end{cases}\\ V=0,043.22,4=0,9632(l)\\ m_{halogen}=m_{Br_2}+m_{I_2}=0,04.160+0,003.254=7,162(g)\\ b,\\ NaCl+AgNO_3\to AgCl+NaNO_3\\ \xrightarrow{\text{Từ PTHH}} \ n_{AgCl}=n_{NaCl}=0,086(mol)\\ m_{AgCl}=0,086.143,5=12,341(g)\\ c,\\ 2KMnO_4+16HCl\to 2KCl+2MnCl_2+8H_2O+5Cl_2\\ \xrightarrow{\text{Từ PTHH }}\ n_{KMnO_4}=\dfrac{2}{5}.n_{Cl_2}=0,0172(mol)\\ m_{KMnO_4}=0,0172.158=2,7176(g)$