Answer
$H_{2}(g)+2ICl(g) \rightarrow I_{2}(g) + 2HCl(g)$
Work Step by Step
Cancel out the intermediate (HI) and then add up the remaining molecules (There was one $H_{2}$ molecule and $2 ICl$ molecules to the left of the arrow. There was 1 $I_{2}$ molecule and $2 HCl$ molecules to the right of the arrow).
$H_2 +ICl\rightarrow HI +HCl$
$HI+ICl\rightarrow I_2+HCl$
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$H_2+2ICl\rightarrow I_2+2HCl$
