At STP- determine the amount of oxygen in liters that can be prepared from the decomposition of 212 grams of sodium chlorate-1 mol-106g-11-222-467-244-878-4

Question

Toppr Admin · Accepted Answer

The molar mass of sodium chlorate is 106 g/mol.
Number of moles of sodium chlorate $= \frac{212 g}{106 g / m o l} = 2 m o l$
The decomposition reaction of sodium chlorate is as shown.
$2 N a C l O_{3} \to 2 N a C l + 3 O_{2}$
The decomposition of 2 moles of sodium chlorate gives 3 moles of oxygen.
1 mole of oxygen at STP occupies a volume of 22.4 L.
3 moles of oxygen at STP will occupy a volume of $3 \times 22.4 = 67.2 L$ .

At STP, determine the amount of oxygen in liters that can be prepared from the decomposition of 212 grams of sodium chlorate(1 mol=106g).11.222.467.244.878.4

At STP, determine the amount of oxygen in liters that can be prepared from the decomposition of $212$ grams of sodium chlorate( $1$ mol $= 106$ g).
$11.2$
$22.4$
$67.2$
$44.8$
$78.4$