Question

Q. how many grams of oxygen gas is essentially required for complete combustion of 3 mole of butane gas. plz answer this question​

Answer 1

Answer:

312 grams

Explanation:

Mass of oxygen required = 19.5 moles × 16 g/mol = 312 g. Hence, 312 grams of oxygen is essentially required for complete combustion of 3 moles of butane gas.

Answer 2

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The combustion of butane is given by,

C4H10+ 13/2o2 → 4co2 +5h2o

Hence, For combustion of 1 mole of butane 6.5 moles of oxygen is required. Therefore, For 3 moles of butane 19.5 moles of oxygen will be required,

Mass of oxygen required = 19.5mol×16g/mol=312g

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Therefore, answer will be 312g

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