How many moles of Calcium Carbonate are present in 1.75 kg of Calcium Carbonate? (Ar of Ca = 40, Ar of C = 12, Ar of O = 16)
Correct answer: D. 17.5 mol
- A. 1.75 mol
- B. 1750 mol
- C. 0.0175 mol
- D. 17.5 mol
Explanation
The molar mass of CaCO3 (40 + 12 + 16 + 16 + 16) is 100g per mole. If one mole equals 100g, than 1750g (1.75 kg x 1000) equals 1750/100, which is 17.5 mol so, D is the answer.
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About Fundamental Concepts of Chemistry
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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