The volume of oxygen at NTP evolved when 1.70 g of sodium nitrate is heated to a constant mass is:

0.112 litre

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0224 litre

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22.4 litre

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11.2 litre

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Solution

The correct option is B 0224 litre
The chemical equation is:

$2 N a N O_{3} (s) \to 2 N a N O_{2} (s) + O_{2} (g)$

2 moles of NaNO $_{3}$ give one mole of Oxygen/22.4 L oxygen at STP

weight of NaNO $_{3}$ = 85g

2 moles of NaNO $_{3}$ weigh = 2(23+14+48) = 170g

170 g of NaNO $_{3}$ will produce 22.4 L Oxygen at STP

1.70 g will produce $= \frac{1.70}{170} \times 22.4 = 0.224$ litres

Hence, the correct option is $B$

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Similar questions

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)
NH₄NO₃ $\overset{}{\to}$ N₂O + 2H₂O
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam?
(Relative molecular mass of ammonium nitrate is 80.)
(iii) Determine the percentage of oxygen in the ammonium nitrate. (O = 16)