# IF 1.5 MOLES OF OXYGEN REACTS WITH SULPHUR (IV) OXIDE, CALCULATE THE MASS OF TETRAOXOSULPHATE(VI) ACID PRODUCED

To calculate the mass of tetraoxosulfate (VI) acid produced when 1.5 moles of oxygen reacts with sulfur (IV) oxide, we need to determine the balanced chemical equation for the reaction first. The balanced equation is as follows:

**2SO2 + O2 + 2H2O→ 2H2SO4**

From the equation, we can see that 1 mole of oxygen reacts with 2 moles of sulfur(IV) oxide to produce 2 moles of tetraoxosulfate (VI) acid (H2SO4).

Given that we have 1.5 moles of oxygen, we can use stoichiometry to find the corresponding amount of tetraoxosulfate (VI) acid produced.

Using the ratio from the balanced equation, we have:

**1 mole O2 : 2 moles H2SO4**

**1.5 moles O2 : x moles H2SO4**

By cross-multiplying and solving for x, we find:

**x = (1.5 moles O2 * 2 moles H2SO4) / 1 mole O2 = 3 moles H2SO4**

Now, to calculate the mass of tetraoxosulfate (VI) acid produced, we need to multiply the number of moles by its molar mass.

The molar mass of H2SO4 is calculated as follows:

**(2 atomic mass of H) + atomic mass of S + (4 atomic mass of O) = 2 (1.0 g/mol) + 32.0 g/mol + 4 (16.0 g/mol) = 98.0 g/mol**

Therefore, the mass of tetraoxosulfate (VI) acid produced can be calculated as:

**Mass = number of moles * molar mass**

**= 3 moles * 98.0 g/mol**

**= 294.0 g**

Hence, the mass of tetraoxosulfate (VI) acid produced when 1.5 moles of oxygen reacts with sulfur(IV) oxide is 294.0 grams.