**Question**

**SO _{3} is prepared by oxidizing SO_{2} with oxygen. How many grams of SO_{2} will be required to produce 40 g of SO_{3} ? [O = 16, S = 32] [Ans. 32 g] **

**Answer**

The balanced chemical equation for the reaction of SO_{2} with O_{2} to form SO_{3} is:

2 SO_{2} + O_{2} → 2 SO_{3}

Molar mass of SO_{3} = 1 × 32 + 3 × 16 = 80

Number of moles of SO_{3} = 40\over 80 = 0.5 mol

Since the molar ratio of SO_{2} to SO_{3} is 1:1, we need the same number of moles of SO_{2} as SO_{3} to react completely. Therefore, the number of moles of SO_{2} required is also 0.5 mol.

Finally, we can use the molar mass of SO_{2} to calculate the mass of SO_{2} required:

Mass of SO_{2} = 0.5 × 64 = 32 g

Therefore, 32 g of SO_{2} will be required to produce 40 g of SO_{3}.