(B) When $KI$ reacts with concentrated $H_2SO_4$,it first produces $HI$ (hydrogen iodide). Since $HI$ is a strong reducing agent and concentrated $H_2

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(B) When $KI$ reacts with concentrated $H_2SO_4$,it first produces $HI$ (hydrogen iodide). Since $HI$ is a strong reducing agent and concentrated $H_2SO_4$ is an oxidizing agent,the $HI$ is further oxidized to $I_2$ (iodine). The reaction is: $2KI + 2H_2SO_4 \rightarrow K_2SO_4 + 2H_2O + I_2 + SO_2$.