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Calculate the percentage of oxygen in su...

Calculate the percentage of oxygen in sulphuric acid, `H_(2) SO_(4)`

A

`82.18%`

B

`3.69%`

C

`65.31%`

D

`50%`

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The correct Answer is:
To calculate the percentage of oxygen in sulfuric acid (H₂SO₄), we will follow these steps: ### Step 1: Determine the molar mass of H₂SO₄ - **Hydrogen (H)**: There are 2 hydrogen atoms, and the atomic mass of hydrogen is approximately 1 g/mol. \[ \text{Mass of H} = 2 \times 1 = 2 \text{ g/mol} \] - **Sulfur (S)**: There is 1 sulfur atom, and the atomic mass of sulfur is approximately 32 g/mol. \[ \text{Mass of S} = 1 \times 32 = 32 \text{ g/mol} \] - **Oxygen (O)**: There are 4 oxygen atoms, and the atomic mass of oxygen is approximately 16 g/mol. \[ \text{Mass of O} = 4 \times 16 = 64 \text{ g/mol} \] Now, we can calculate the total molar mass of H₂SO₄: \[ \text{Molar mass of H₂SO₄} = \text{Mass of H} + \text{Mass of S} + \text{Mass of O} = 2 + 32 + 64 = 98 \text{ g/mol} \] ### Step 2: Calculate the mass of oxygen in H₂SO₄ From our previous calculation, we found that the total mass of oxygen in H₂SO₄ is: \[ \text{Mass of O} = 64 \text{ g/mol} \] ### Step 3: Calculate the percentage of oxygen in H₂SO₄ To find the percentage of oxygen in sulfuric acid, we use the formula: \[ \text{Percentage of O} = \left( \frac{\text{Mass of O}}{\text{Molar mass of H₂SO₄}} \right) \times 100 \] Substituting the values we calculated: \[ \text{Percentage of O} = \left( \frac{64}{98} \right) \times 100 \approx 65.31\% \] ### Final Answer The percentage of oxygen in sulfuric acid (H₂SO₄) is approximately **65.31%**. ---

To calculate the percentage of oxygen in sulfuric acid (H₂SO₄), we will follow these steps: ### Step 1: Determine the molar mass of H₂SO₄ - **Hydrogen (H)**: There are 2 hydrogen atoms, and the atomic mass of hydrogen is approximately 1 g/mol. \[ \text{Mass of H} = 2 \times 1 = 2 \text{ g/mol} \] - **Sulfur (S)**: There is 1 sulfur atom, and the atomic mass of sulfur is approximately 32 g/mol. ...
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Oleum or fuming sulphuric acid contains SO_(3) dissolved in sulphuric acid and has the molecular formula H_(2)S_(2)O_(7) , It is formed by passing SO_(3) in H_(2)SO_(4) . When water is added to oleum, SO_(3) reacts with water to form H_(2)SO_(4) . SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq) As a result, mass of H_(2)SO_(4) increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling of oleum. % Labelling of oleum = Total mass of H_(2)SO_(4) present in oleum after dilution or = Mass of H_(2)SO_(4) initially present + Mass of H_(2)SO_(4) produced after dilution From this, the percentage composition of H_(2)SO_(4) and SO_(3) (free) and SO_(3) (combined) can be calculated. The percentage of free SO_(3) and H_(2)SO_(4) in 112% H_(2)SO_(4) is

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