The mass of oxygen gas obtained by the decompostion of 57.93 g of silver oxide is:
(Molar mass of $A g_{2} O = 231.735 {g mol}^{- 1}$ )

16 g

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4 g

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8 g

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32 g

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Solution

The correct option is B 4 g
The unbalanced eqauation is,
$A g_{2} O \to A g + O_{2}$
1 mol of $A g_{2} O$ contains 1 mol of $O$ atoms;
1 mol of $O_{2}$ contains 2 mol of $O$ atoms.
Applying POAC on $O$ atom,
$1 \times mol of$ $A g_{2} O = 2 \times mole of$ $O_{2}$
Moles of $O_{2}$ = $0.5 \times moles of$ $A g_{2} O$
So, we can calculate the moles of $A g_{2} O$ , convert it into the moles of $O_{2}$ and finally to grams of $O_{2}$ .
Moles of $A g_{2} O$ = $\frac{57.93}{231.73}$ = 0.25 mol
Moles of $O_{2}$ = 0.125 mol
Mass of $O_{2}$ = $0.125 \times 32$ = 4 g

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