Consider the balanced chemical equation for the combustion of methane (CH4).
2 answers:
<h3>Answer:</h3>
89.6 L of O ₂
<h3>Solution:</h3>
The balanced chemical equation is as,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,
44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂
So,
88 g CO₂ will be produced by = X L of O₂
Solving for X ,
X = (88 g × 44.8 L) ÷ 44 g
X = 89.6 L of O ₂
Took a guess and got it right! D) 89.6 L
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