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Volume of Gases and Number of Moles

The amount of...

Question

The amount of $M g$ required to replace $A g$ from 500 ml of 0.5 M $A g N O_{3}$ solution is:

1.2 g

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2.4 g

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3 g

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6 g

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Solution

The correct option is C 3 g
$M g + 2 A g N O_{3} = 2 A g + M g (N O_{3})_{2}$
500 ml of 0.5 M $A g N O_{3} = \frac{500 m L}{1000 m L / L} \times 0.5 M = 0.25$ moles
0.25 moles $A g N O_{3} = \frac{0.25}{2} = 0.125$ moles Mg
The atomic mass of $M g = 24.3$ g/mol.
Mass of $M g = 24.3 \times 0.125 = 3.0$ g

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