What is the volume (in dm3) of oxygen at STP required for complete combustion of 32 g of CH4? (mol. Wt of CH4=16)
Correct answer: A. 89.6.
- A. 89.6.
- B. 98.4.
- C. 189.6.
- D. 169.5
Explanation
The correct option is AEquation for the above reaction: CH4 + 2O2 → CO2 + 2H2Omolecular mass of CH4 = 16mass of CH4 given = 32 gno of moles of CH4: 32 / 16 = 2 molesstoichiometric ratio of CH4 : O2 = 1 : 2no of moles of oxygen used: 2 x 2 = 4 molesvolume of 1 mole of gas at s.t.p = 22.4 dm3volume of oxygen used = 4 x 22.4 = 89.6 dm3
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