Zinc reacts with $C u {S O}_{4}$ according top the equation $Z n + C u {S O}_{4} ⟶ Z n {S O}_{4} + C u$ . if excess of zinc is added to $100 m l$ of $0.05 M$ $C u {S O}_{4}$ the amount of copper formed is:

0.6354 g

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0.317 g

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3.177 g

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6.354 g

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Solution

The correct option is A $0.317 g$
$1$ mole of $C u S O_{4}$ gives $1$ mole of copper.
Given moles of $C u S O_{4} = 100 \times 0.05 \times 10^{- 3} = 5 \times 10^{- 3}$
Moles of copper formed $= 5 \times 10^{- 3}$
$∴$ Amount of copper formed $= 63.54 \times 5 \times 10^{- 3} = 0.3177 g$

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