=(−787.00−857.52)+84.70=−1559.82 

) PTHH + xác định 2KMnO4+16HCl→2KCl+2MnCl2+5Cl2+8H2O2KMnO_4 + 16HCl 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O2KMnO4​+16HCl→2KCl+2MnCl2​+5Cl2​+8H2​O Chất oxi hóa: KMnO₄ (Mn⁷⁺ → Mn²⁺) Chất khử: HCl (Cl⁻ → Cl₂) Quá trình oxi hóa: 2Cl−→Cl2+2e2C Cl_2 + 2e2Cl−→Cl2​+2e Quá trình khử: MnO4−→Mn2+MnO_4^- \rightarrow Mn^{2+}MnO4−​→Mn2+ b) Tính khối lượng KMnO₄ nNaI=0.2×0.1=0.02 moln_{NaI} = 0.2 \times 0.1 = 0.02 \, molnNaI​=0.2×0.1=0.02mol PTHH: Cl2+2NaI→2NaCl+I2Cl_2 + 2NaI \rightarrow 2NaCl + I_2Cl2​+2NaI→2NaCl+I2​ → nCl2=0.01 moln_{Cl_2} = 0.01 \, molnCl2​​=0.01mol C= 0,632 g

v=Δt/ΔC​=40.22/0.10​=0.03M/s V= 0,03 M/