Labeling of Oleum & $H_2O_2$
Stoichiometry | Concentration Terms | Class 11 Chemistry
1. Labeling of Oleum
Oleum (Fuming Sulphuric Acid) is a mixture of $H_2SO_4$ and free $SO_3$ gas. It is represented as $H_2S_2O_7$. Oleum is labeled as a percentage greater than 100% (e.g., 109% Oleum).
What does "109% Oleum" mean?
It means that if we take 100g of this Oleum sample and add water, the free $SO_3$ reacts with water to form $H_2SO_4$. The total mass of pure $H_2SO_4$ obtained will be 109g.
It means that if we take 100g of this Oleum sample and add water, the free $SO_3$ reacts with water to form $H_2SO_4$. The total mass of pure $H_2SO_4$ obtained will be 109g.
Logic & Calculation
The extra percentage ($x - 100$) represents the mass of water added.
$$ SO_3 + H_2O \rightarrow H_2SO_4 $$Stoichiometry: 80g $SO_3$ reacts with 18g $H_2O$.
$$ \% \text{ Free } SO_3 = \frac{80 \times (x - 100)}{18} $$
Where $x$ is the label percentage (e.g., 109).
2. Volume Strength of $H_2O_2$
Hydrogen Peroxide solution is labeled as '10V', '20V', etc. This refers to the volume of Oxygen gas evolved at STP upon decomposition.
Definition: "20 Volume $H_2O_2$" means that 1 Liter of this solution will produce 20 Liters of $O_2$ gas at STP.
Stoichiometry
$$ 2H_2O_2 \rightarrow 2H_2O + O_2 $$2 moles of $H_2O_2$ produce 1 mole (22.4 L at STP) of $O_2$.
Important Relations (Shortcuts)
If 'V' is the Volume Strength:
$$ \text{Volume Strength} = 11.2 \times \text{Molarity (M)} $$
$$ \text{Volume Strength} = 5.6 \times \text{Normality (N)} $$
Strength in g/L: $S = N \times \text{Eq. Wt} = N \times 17$.
Practice Quiz
Test your numerical skills on Oleum and Peroxide.
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