How many grams of oxygen will be formed by the action of 12 g of sodium peroxide on water? Calculate also the volume of the gas at NTP.

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Solution

$2 N a_{2} O_{2} + 2 H_{2} O ⟶ O_{2} + 4 N a O H$
$2 \times 78 g$ $32 g$
$156 g$ of $N a_{2} O_{2}$ produces $32 g$ of $O_{2},$
$12 g$ of $N a_{2} O_{2}$ produces $= \frac{32 \times 12}{156} = 10.66 g .$
Density of $O_{2}$ at NTP $= 1.428 g / m L .$
$\frac{M a s s}{D e n s i t y} = V o l .$
$\frac{10.66}{1.428} = 7.46 m L$
Vol. of $O_{2}$ at NTP is $7.46 m L .$

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