Question

When sodium bicarbonate is heated 1.806 x 1024
molecules of water is obtained. Then find the vol-
ume of CO.(g) obtained at NTP and amount of
NaHCO, needed for this reaction.
2NaHCO3(s) - Na2CO3(s) + H2O(g) + CO2(9)
(A) 67.2L, 504 g (B) 22.4L, 252 g
(C) 67.2 L, 252 g
(D) 22.4 L, 5049​

Answer 1

Answer:

hope this answer helps u

Answer 2

Option (D) is the correct answer

The volume of CO(g) obtained at NTP and the amount of NaHCO3, needed for this reaction is 504 gm

Explanation:

Given When sodium bicarbonate is heated $1.806$ x$10^{24}$ molecules of water is obtained.

the volume of CO obtained at NTP and the amount of NaHCO3 is to be determined.

$2NaHCO3(s) --- Na2CO3(s) + H2O(g) + CO2(g)$

Here Sodium bicarbonate gives sodium carbonate on decomposition.

$2NaHCO3 ----- Na2CO3$

n moles of H2O =$1.8$× $10^{24}$/$6.023$×$10^{23}$  

=2.9

n moles of CO = 3 moles

Here 3 moles of NaHCO3 = total amount of NaHCO3

= 3×84 gm of  NaHCO3

= 504 gms

504 gms of Sodium bicarbonate at 22.4 litre STP

The volume of CO(g) obtained at NTP and the amount of NaHCO3, needed for this reaction is 504 gm

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