The volume of O₂ : 21 L
<h3>Further explanation</h3>
Given
8.7 grams of C₂H₄
Required
Volume O₂
Solution
Reaction
C₂H₄ + 3 O₂ ⇒ 2 CO₂ + 2 H₂O
mol C₂H₄(MW= 28 g/mol) :
= mass : MW
= 8.7 g : 28 g/mol
= 0.311
From the equation, mol O₂ :
= 3/1 x mol C₂H₄
= 3/1 x 0.311
= 0.933
At STP, 1 mol gas=22.4 L, so for 0.933 mol :
= 0.933 x 22.4 L
= 20.899 L ≈ 21 L
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