An ammeter reading shows a current of 0.50 A running through the iron hull. How many grams of $F e (s)$ are lost in 1.0 hour?

261 mg

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521 mg

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1041 mg

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2084 mg

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Solution

The correct option is B 521 mg
The quantity of electricity passed $= Q (C) = 0.5 A \times 3600 s = 1800 C$ .
Molar mass of iron = 55.8 $\frac{g}{m o l}$
Moles of electrons passed $= \frac{Q (C)}{96500 \frac{C}{m o l e^{-}}} = \frac{1800 C}{96500 \frac{C}{m o l e^{-}}} = 0.01865 m o l e^{-}$
The weight of iron lost $= m o l a r m a s s o f F e \times m o l e r a t i o \times m o l e s o f e l e c t r o n s p a s s e d = 55.8 \frac{g}{m o l} \times \frac{1 m o l F e}{2 m o l e^{-}} \times 0.01865 m o l e^{-} = 0.521 g = 521 m g .$

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