Question

32g of SO2 reacts with 48g of O2 to produce SO3. Calculate and find out the limiting reagent in the manufacture of SO3 .(Given molar mass of SO2 = 64 g mol^-1 and O2 32g mol^-1)

Answer 1

So2 is limiting reactant because
1 mole of s02 combines with 1 mole of O2
So 0.5 mole of s02 combine with 0.5 mole of O2
So so2 is a limiting reactant

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Answer 2

32g of SO₂ reacts with 48g of O₂ to produce SO₃.

we have to find out the limiting reagent in the manufacture of SO₃.

Given,

• mass of SO₂ = 32 g
• mass of O₂ = 48 g
• molar mass of SO₂ = 64 g/mol
• molar mass of O₂ = 32g/mol

no of moles of a substance = mass of given substance/molar mass of that substance

so, no of moles of SO₂ = mass of SO₂/molar mass of SO₂

= 32/64 = 0.5 mol

similarly no of moles of O₂ = mass of O₂/molar mass of O₂

= 48/32 = 1.5 mol

chemical reaction between SO₂ and O₂ is given by,

$\quad 2SO_2+O_2\rightarrow 2SO_3$

we see, 2 moles of SO₂ react with one mole of O₂ and produce 2 moles of SO₃.

∴ 0.5 mol of SO₂ will react with 0.25 mol of O₂ and will produce 0.5 mol of SO₃.

we see, only 0.25 mol out of 1.5 mol of O₂ will react when 0.5 mol of SO₂ (complete amount of SO₂) reacts, after that, no further reaction occurs because shortage of SO₂.

hence, SO₂ is limiting reagent.

Therefore SO₂ is limiting reagent in the reaction.

as SO₂ is limiting reagent, reaction occurs on the basis of it.

∵ 2 mol of SO₂ produces 2 mol of SO₃.

∴ 0.5 mol of SO₂ will produce 0.5 mol of SO₃.

so the mass of SO₃ = no of moles of SO₃ × Molar mass of SO₃.

= 0.5 mol × 80 g/mol

= 40g

Therefore the mass of SO₂ formed after the reaction is 40g.