C + O2 ---> CO2
1000 g C / 12 g/mole = 83.33 moles C
83.33 moles O2 needed.
V = nRT/P = 83.33 * 0.08206 * 273 / 1 = 1867 L O2
Air is 21% oxygen, so 1866 / 0.21 = 8890 L air
2007-05-16 01:26:28
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answer #1
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answered by TheOnlyBeldin 7
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No of moles of Carbon = wt/atomic wt = 1000/12 =83.33=250/3
according to equation
C+O2---->CO2, we need the same number of moles of oxygen.
also, 1 mole of any gas at STP occupies 22.4L
so O2 will occupy = (250/3)X22.4 = 1866.67 L
Also since air has 20% O2, so to get the above mentioned amount of O2, volume of air needed will be:
5X1866.67 = 9333.33L
2007-05-16 01:23:22
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answer #2
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answered by alien 4
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at STP, C+O2 -----> CO2
so for 12gms of C (atomic mass), 32 gms (molecular mass) of O2 is required. so for 1kg of carbon, 2666.67gms (2.67kg) of O2 is needed.
concentration of O2 in air is-21%, so, air needed is-12698.4gm (12.69kg)
PS-:if the density of air is known, then the volume required can be calculated
2007-05-16 01:22:05
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answer #3
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answered by mantu1411 2
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n= volume divinded by molar volume
2007-05-16 01:26:41
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answer #4
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answered by Nikki S 1
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5 sacks of sugar....hehe
2007-05-16 01:13:48
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answer #5
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answered by Zen-Psycholgy Guy 2
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