Question

Oleum sample is identified as 102.25%. The % free so3 present in this sample is

Answer 1

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

Answer 2

Given:

$\[102.25\% \]$ Oleum sample

To Find: $\[\% \]$ free $\[S{O_3}\]$

Solution:

The formula for Oleum is $\[{H_2}S{O_4} \cdot S{O_3}\]$.

When water is added to Oleum, it forms Sulphuric acid.

Water reacts with $\[S{O_3}\]$ to give sulphuric acid.

$\[S{O_3} + {H_2}O \to {H_2}S{O_4}\]$

$\[102.25\% \]$ Oleum means $\[100\,g\]$ of Oleum react with water to prepare $\[102.25\,g \]$ of Sulphuric acid.

Thus, the mass of water required $\[ = 102.25\,g - 100\,g = 2.25\,g\]$

From the above reaction, it is clear that one mole of $\[S{O_3}\]$ reacts with one mole of water to form one mole of sulphuric acid.

The molar masses of these compounds are:

$\[\begin{gathered} S{O_3} = 32 + 3 \times 16 \hfill \\ S{O_3} = 80\,g/mol \hfill \\ \end{gathered} \]$

$\[\begin{gathered} {H_2}O = 1 \times 2 + 16 \hfill \\ {H_2}O = 16\,g/mol \hfill \\ \end{gathered} \]$

$\[\begin{gathered} {H_2}S{O_4} = 2 \times 1 + 32 + 4 \times 16 \hfill \\ {H_2}S{O_4} = 98\,g/mol \hfill \\ \end{gathered} \]$

Thus, $\[80\,g\]$ of $\[S{O_3}\]$ reacts with $\[18\,g\]$ of water to form $\[98\,g\]$ of sulphuric acid.

Therefore, $\[2.25\,g\]$ water reacts with $\[\frac{{80}}{{18}} \times 2.25\,\,g = 10\,g\]$ of $\[S{O_3}\]$.

Therefore, the percentage of $\[S{O_3}\]$ is $\[10\% \]$.

Hence, $\[\% \]$ free $\[S{O_3}\]$ in the sample is $\[10\% \]$.

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