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(C) The overall reaction can be obtained by adding the two given equations:$(A + 2B 	o I) + (I + C 	o B + D)$Simplifying,we get: $A + B + C 	o D$Si

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$5$

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(C) The overall reaction can be obtained by adding the two given equations:$(A + 2B 	o I) + (I + C 	o B + D)$Simplifying,we get: $A + B + C 	o D$Since $C$ is in excess,the limiting reagent is determined by $A$ and $B$.Initially,we have $5 \, 	ext{moles}$ of $A$ and $6 \, 	ext{moles}$ of $B$.According to the net reaction $A + B + C 	o D$,$1 \, 	ext{mole}$ of $A$ reacts with $1 \, 	ext{mole}$ of $B$ to produce $1 \, 	ext{mole}$ of $D$.Since $A$ is the limiting reagent $(5 \, 	ext{moles} Therefore,$5 \, 	ext{moles}$ of $A$ will react with $5 \, 	ext{moles}$ of $B$ to produce $5 \, 	ext{moles}$ of $D$.

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