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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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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