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9g water is added into 100g of oleum sample labelled as 112% H2SO4, then the amount of free SO3 remaining in the solution is? (Working pls)

Vikas TU
14149 Points
4 years ago
The condition for response is :

SO3+H2O— — −>H2SO4

This infers the measure of moles of water responds with same number of moles of SO3 in the specimen :

The quantity of moles of water included is = MASS/MOLECULAR MASS =9/18= 0.5 mole of water

This responds with an equivalent number of moles of SO3 :

Unique number of moles of SO3 in the specimen is = mass/atomic mass

Mass is computed as:

Be that as it may, the quantity of moles of H2S04 framed would be : y/80

The aggregate mass of H2SO4 would be mole X atomic mass = 98y/80

Mass of H2 SO4 in 100 gm would be 100-y

Be that as it may, new mass framed + existing mass must be equivalent to 109.9% henceforth ;

98y/80 + (100-y) = 100.9

Y= 53.33 gm

Number of moles is 53.33/80

= 0.6667

Number of moles staying after response with water :

= 0.6667-0.5

=0.1667

The mass that remaining parts is = moles X sub-atomic mass

= 0.1667 X 80

= 13.33 gm