Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: peterpan1372 on September 07, 2017, 05:38:16 PM
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Hello,
can someone tell me how I can solve the question beneath in the fastest way possible? (for the BMAT)
Following equation:
CH_4 + 2O_2 -> CO_2 +2H_2O
If 1.6 g of methane is completely burned in 8 g of oxygen(an excess) to produce 4,4 g of CO2, what mass of oxygen is left unreacted?
So we have: 32g O2 -> 44g CO2 and 36g H2O,
However, as you might think of yourself, I have difficulties with solvin such questions, which is why I need your help please..
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Classic limiting reagent problem. As you are already told oxygen is in excess all you need to do is to find out how many grams of oxygen will react with methane.
So we have: 32g O2 -> 44g CO2 and 36g H2O
Not 32 g of oxygen, check it over.
But in general you are on the right track, just add methane to these numbers.
(note that 4.4 g is exactly one tenth of 44 g, which makes calculations pretty easy and fast in this case)
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Hello,
so we have 64g of oxygen, you're right.
So: 16g of Methane react with 64g of Oxygen, to form 44g of CO2 and 36g of Water. However, how do I have to go on from here?
Further, we have 1/10 mole of methane reacting with 1/8 mole of oxygen, giving 1/10 mole of CO2.
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my fqult ... forget everthing. I have calculated that there will be 4.8g of oxygen left since methane is the limiting factor, so we have 3.2g lf oxzygen used up, so:8-3.2=4.8-3.2=1.6
However, oxygen still reachts with H2, so shouldnt that be taken into consideration,too?
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my fqult ... forget everthing. I have calculated that there will be 4.8g of oxygen left since methane is the limiting factor, so we have 3.2g lf oxzygen used up, so:8-3.2=4.8-3.2=1.6
No, that's not correct. Looks to me like the same mistake you did earlier - 16 g of methane reacts with 64 g of oxygen, not 32 g of oxygen.
However, oxygen still reachts with H2, so shouldnt that be taken into consideration,too?
It is already taken into consideration, oxygen doesn't react separately with methane and separately with hydrogen.
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Hello, but in the question it says that 1.6g of Methane react with 8g of Oxygen. we have 2x oxyggen, so 2x1.6=3.2 g of oxygen is used up....
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No. Do you know how to read the reaction equation?
http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
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Could you then tell me finallybwhat the right way is?
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Try to follow the explanation given at the link.
You have 1.6 g of methane - how many moles is that?
With how many moles of oxygen will it react?
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tis one tenth mole of methane. So 6.4g of oxygen will react leaving 1.6g left, or?
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tis one tenth mole of methane. So 6.4g of oxygen will react leaving 1.6g left, or?
Yes.
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It is a simple question if you convert everything in moles.
We know that 1 mole of methane reacts with 2 moles of oxygen to give 1 mole of CO2 and 2 moles Water
CH4 + 2 O2 :rarrow: CO2 + 2 H2O
Given 1.6 g of methane = 0.1 mole of methane
which should react with 0.2 moles of O2.
And 0.2 moles of O2 = 3.2 g of O2
Thus 8g-3.2g= 4.8g of oxygen is unreacted
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Thus 8g-3.2g= 4.8g of oxygen is unreacted
Please read the forum rules (http://www.chemicalforums.com/index.php?topic=65859.0) - giving final answers is strictly prohibited.
Besides, the answer you gave is wrong:
And 0.2 moles of O2 = 3.2 g of O2
Think it over.
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tis one tenth mole of methane. So 6.4g of oxygen will react leaving 1.6g left, or?
yes that is right ...
Always read equations in moles / or number of molecules or volumes if gases are there at the same temperature and pressure.Numbers in balanced equations do not represent grams ...they are for moles or molecules or atoms ...