What volume of oxygen, measured at 27.0°C and 99.50 kPa, is needed for the combustion of 1.03 kg of coal?
What volume of oxygen, measured at 27.0°C and 99.50 kPa, is needed for the combustion of 1.03 kg of coal?
(For this problem, assume coal is 100% carbon.)
C(s) + O2(g) CO2(g)
the answer must be in liters of oxygen gas.
Please just give me the straight up answer!
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Verified answer
C + O2 = CO2
moles C = 1030/ 12 g/mol=85.8 = moles O2
99.50 kJ= 0.982 atm
T = 300 K
V = 85.8 x 0.08206 x 300/ 0.982=2151 L
No of moles of Carbon = wt/atomic wt = 1000/12 =eighty 3.33=250/3 in accordance to equation C+O2---->CO2, we prefer a similar form of moles of oxygen. additionally, a million mole of any gas at STP occupies 22.4L so O2 will occupy = (250/3)X22.4 = 1866.sixty seven L additionally because of fact that air has 20% O2, so which you need to get the above pronounced volume of O2, volume of air mandatory would be: 5X1866.sixty seven = 9333.33L