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5 answers

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 · answer #1 · answered by TheOnlyBeldin 7 · 1 0

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 · answer #2 · answered by alien 4 · 0 0

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 · answer #3 · answered by mantu1411 2 · 0 0

n= volume divinded by molar volume

2007-05-16 01:26:41 · answer #4 · answered by Nikki S 1 · 0 0

5 sacks of sugar....hehe

2007-05-16 01:13:48 · answer #5 · answered by Zen-Psycholgy Guy 2 · 0 0

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