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What is the mass of free SO3 in 50g of oleum when it is labelled 100.9%?

What is the mass of free SO3 in 50g of oleum when it is labelled 100.9%?

Grade:11

2 Answers

Vikas TU
14149 Points
6 years ago
SO3+H2O———−>H2SO4
If we have y gm of SO3 and (100-y) gm of H2SO4 in 100gm of Oleum then
Oleum would react with water to form sulphuric (IV) acid hence the number of moles of oleum taking part in the reaction to molecular mass would be            :     y/80
But  the number of moles of H2S04  formed would be         : y/80
 The total mass of H2SO4 would be                           mole X molecular mass  = 98y/80
Mass of H2 SO4 in 100 gm would be  100-y
But new mass formed +  existing mass must be equal to 109.9% hence ;
 98y/80  + (100-y)  = 100.9
Y =4
Mass in 50gm og oleum is 4/2 =2gm
Rajkamal Prasad
11 Points
one year ago
Since, Oleum = H2SO4.SO3
Oleum + water means
                SO3 + H2O\rightarrow H2SO4
100.9°/° oleum means 100 gm oleum and
0.9 gm water react with free SO3
In this question we have given 50 gm oleum that mean 0.9/2 gm water will be recquired.
 
SO3  +    H2O   \rightarrow  H2SO4
           (0.45gm)
Using stoichiometry we get
 1mole SO3 requires 1 mole H2O
        1mole SO3 =   1*   given mass
   molar mass (H2O)
1 mole SO3 =      0.45    =    1
         18            40
          mass of SO3            =            1   molar mass of SO3 (80gm)          40
                mass of SO3    =   80   =   2
    40
Mass of free SO3 in 50 gm is 2gm..
 
 

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