The volume occupied by 8 gram of O2 at STP
Correct answer: C. 5.6dm³
- A. 11.2dm³
- B. 22.4dm³
- C. 5.6dm³
- D. None of these
Explanation
Moles = Mass / Mr = 8/32 = 0.25 mol of O2.Volume= Moles x 22.4dm³(at STP) = 0.25 x 22.4 = 5.6dm³
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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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