9g water is added into 100g oleum sample labelled as 112 percent h2so4 then the amount of free so3 remaning in the solution is
a.14.93L at STP
b. 7.46L at STP
c.3.78L at STP
d.11.2L at STP
Dear Student,
112% oleum means 100g is the sample and 12g is the water required by sample to completely combine SO3 in the sample.
And according to reaction
SO3 + H2O = H2SO4
80g 18g 98g
So, 12g water combines with 53.3g SO3 and hence in 100g sample 53.3g is SO3 and 46.7g is H2SO4.
Now, If another 9g of water is added to it, another 40g of SO3 is converted to H2SO4.
So, remaining free SO3 is 15.3g = 0.2 moles
At STP, 1 mol has volume 22.4 L. SO, 0.2 moles has volume = 3.78 L.
So, the correct option is C.
Regards
112% oleum means 100g is the sample and 12g is the water required by sample to completely combine SO3 in the sample.
And according to reaction
SO3 + H2O = H2SO4
80g 18g 98g
So, 12g water combines with 53.3g SO3 and hence in 100g sample 53.3g is SO3 and 46.7g is H2SO4.
Now, If another 9g of water is added to it, another 40g of SO3 is converted to H2SO4.
So, remaining free SO3 is 15.3g = 0.2 moles
At STP, 1 mol has volume 22.4 L. SO, 0.2 moles has volume = 3.78 L.
So, the correct option is C.
Regards