See image | AITS & Test Series — Chem-Mantra

Name: JEE Advanced Chemistry
Price: 7999 INR
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Question

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Answer: 00000.33

💡 Solution & Explanation

Let 2 6 2 4 molesof C H and molsof C H x y   Moles of O2 used 7x +3y 2       Moles of O2 available = 130/32 7 130 3 2 32 x y   

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