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1. 100 cm3 of methane, CH4, is completely burned in 400 cm3 of oxygen.

What is the final volume of the gas mixture, in cm3, when all volumes are measured at room temperature and pressure?

A) 100

B) 200

C) 300

D) 400

Please somebody give the answer with detailed working. Thanks. God Bless you.
2. (Original post by dajjal619)
Please somebody give the answer with detailed working. Thanks. God Bless you.
No. We won't just give you a full worked solution, it doesn't help you in the long run.

Start by writing out a balanced chemical equation.

edit: turns out we will do it for you
3. (Original post by dajjal619)
100 cm3 of methane, CH4, is completely burned in 400 cm3 of oxygen.

What is the final volume of the gas mixture, in cm3, when all volumes are measured at room temperature and pressure?

A) 100

B) 200

C) 300

D) 400

Please somebody give the answer with detailed working. Thanks. God Bless you.
moles of methane = 100/24000
moles of oxygen = 400/24000
CH4 + O2 -> CH4O2
methane is the limiting reagent therefore you use the number of moles of methane found then
(100/24000) * 24000 = 100
but then you have 300 O2 spare
so ....
4. (Original post by dajjal619)
100 cm3 of methane, CH4, is completely burned in 400 cm3 of oxygen.

What is the final volume of the gas mixture, in cm3, when all volumes are measured at room temperature and pressure?

A) 100

B) 200

C) 300

D) 400

Please somebody give the answer with detailed working. Thanks. God Bless you.
Its about the molar ratio. First you need to work out the balanced symbol equation: CH4 + 2O2 --> CO2 + 2H2O

This means that 1 mol of CH4 will react with 2 mols of oxygen to form 1 mol of carbon dioxide and 2 mols of water.

So 100cm3 reacts with 200cm3 (as it is a ratio of 1:2) to form 100cm3 of carbon dioxide (we are not interested in the water as it is not a gas)

However the oxygen is in excess, so the answer is the volume of the excess plus the volume of the carbon dioxide:
(400-200)+100 =300 the answer is C
5. Thanks Sir. God Bless you.
6. Yes sir u r right. Trying from our side is necessary. But believe me i got the answer be4 posting the question, I just wanted to confirm . Thanks for your concern.

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