From 64 g oxygen, 11.2L oxygen gas at S.T.P. and 6.022*10^23 oxygen atoms are removed from the oxygen container. Find the mass of the oxygen left.

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Solution

Dear Yukta,

$m o l e s p r e s e n t i n 64 g O x y g e n (O_{2}) = \frac{64}{32} = 2 m o l e s a t S T P 11.2 L O x y g e n (O_{2}) = 0.5 m o l e s a l s o r e m o v e d 6.023 \times 10^{23} a t o m s o f o x y g e n = 3.011 \times 10^{23} m o l e c u l e s o f O x y g e n (O_{2}) w h i c h i s a l s o e q u a l t o 0.5 m o l e s o f o x y g e n (O_{2}) S o t o t a l m o l e s o f o x y g e n (O_{2}) r e m o v e d i s 1 m o l e, m o l e s o f o x y g e n (O_{2}) l e f t i s 2 - 1 = 1 m o l e s i . e . e q u a l t o 32 g o x y g e n (O_{2}) .$

Regards

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Similar questions

One mole of oxygen gas at STP is equal to...........

(a) $6.022 \times 10^{23}$ molecules of oxygen

(b) $6.022 \times 10^{23}$ atoms of oxygen

(d) 32 g of oxygen

12 \times 10^{24}

6 \times 10^{23}

[O = 16]

6.022 \times 10^{23}

which has more weight 6.022*1023 atoms of oxygen or carbon?

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