When 10 g of copper and 10 g of iodine are | vedclass

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(C) Step $1$: Calculate the moles of reactants.Molar mass of $Cu \approx 63.5 \ g/mol$,Molar mass of $I_2 \approx 253.8 \ g/mol$.Moles of $Cu = \frac{

Vedclass · Accepted Answer

$15$

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(C) Step $1$: Calculate the moles of reactants.Molar mass of $Cu \approx 63.5 \ g/mol$,Molar mass of $I_2 \approx 253.8 \ g/mol$.Moles of $Cu = \frac{10}{63.5} \approx 0.157 \ mol$.Moles of $I_2 = \frac{10}{253.8} \approx 0.0394 \ mol$.Step $2$: Identify the limiting reagent.According to the equation $2Cu + I_2 \longrightarrow 2CuI$,$1 \ mol$ of $I_2$ requires $2 \ mol$ of $Cu$.For $0.0394 \ mol$ of $I_2$,we need $2 	imes 0.0394 = 0.0788 \ mol$ of $Cu$.Since we have $0.157 \ mol$ of $Cu$ (which is $> 0.0788 \ mol$),$I_2$ is the limiting reagent.Step $3$: Calculate the yield of $CuI$.Moles of $CuI$ produced $= 2 	imes 	ext{moles of } I_2 = 2 	imes 0.0394 = 0.0788 \ mol$.Molar mass of $CuI = 63.5 + 126.9 = 190.4 \ g/mol$.Mass of $CuI = 0.0788 \ mol 	imes 190.4 \ g/mol \approx 15 \ g$.Thus,the correct option is $C$.

When $10 \ g$ of copper and $10 \ g$ of iodine are mixed,calculate the theoretical yield of $CuI$ according to the equation,$2Cu + I_2 \longrightarrow 2CuI$. (in $g$)

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