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Question: The percent free of \(S{O_3}\) is an Oleum is \(20\% \).Label the sample of Oleum in terms of \({H_2...

The percent free of SO3S{O_3} is an Oleum is 20%20\% .Label the sample of Oleum in terms of H2SO4{H_2}S{O_4}.
A)113.5%
B) 104.5%
C)106.75%
D)120%

Explanation

Solution

Hint : The percentage of Oleum should be taken as 100% and oleum is the mixture of diprotic acid and sulphur trioxide. The diprotic acid molecular mass is 98g.

Complete step by step solution :
Oleum is defined as fuming Sulphuric acid . It is a mixture of Sulphuric acid and free Sulphur trioxide.
Oleum chemical formula can be written as H2SO4.XSO3{H_2}S{O_4}.XS{O_3}. Oleum is used in industrial application and other processes. This is used in the Sulfonation process , Nylon manufacturing , Dye manufacturing and Hydrofluoric acid production.
If oleum is released in the upper atmosphere and exists as sulphuric acid droplets. This sulphuric acid droplet settles in the cloud and Oleum is not biodegradable.
The molecular weight of Sulphur trioxide is 80g
SO3=32+3×16=80gS{O_3} = 32 + 3 \times 16 = 80g
The molecular weight of Water is 18g
H2O=2+16=18g{H_2}O = 2 + 16 = 18g
The molecular weight of Sulphuric acid is 98g
H2SO4=2+32+4×16=98g{H_2}S{O_4} = 2 + 32 + 4 \times 16 = 98g
From the question concluded that 20% is free sulphuric acid
In the above stoichiometric balanced equation 1 mole of sulphuric trioxide reacting with 1 mole of Water and gives fuming sulphuric acid nothing but oleum
But as per the question let consider
x18moles  of  water    reacting  with  x18moles  of  sulphur  trioxide\frac{x}{{18}}moles\;of\;water\;\;reacting\;with\;\frac{x}{{18}}moles\;of\;sulphur\;trioxide
Number  of  moles  SO3=WeightGram  molecular  weightNumber\;of\;moles\;S{O_3} = \frac{{Weight}}{{Gram\;molecular\;weight}}
So Weight of Sulphur trioxide can be written as X18×80\frac{X}{{18}} \times 80
X18×80\frac{X}{{18}} \times 80 =20%= 20\%
X18=2080 X18=14 X=184=4.5%   \frac{X}{{18}} = \frac{{20}}{{80}} \\\ \frac{X}{{18}} = \frac{1}{4} \\\ X = \frac{{18}}{4} = 4.5\% \\\ \\\
So total oleum is considered as 100% and but free Sulphur trioxide is 4.5%
So the label of Oleum in the Percent ofH2SO4{H_2}S{O_4}is 104.5%104.5\%

Hence answer is (B) 104.5%104.5\%

Note : The oleum contains free sulphur trioxide. The given percentage 20% of Sulphur trioxide is equal to 20g of Sulphur trioxide. The sulphur trioxide contains 3 oxygen atoms in its molecular formula.