The total number of protons in 10 g of calcium carbonate are: (NA=6.023 x 10^23)
Correct answer: A. 3.01 x 10^24
- A. 3.01 x 10^24
- B. 30.1.1024
- C. 4.06 x 10^24
- D. 3.01 x 10^23
Explanation
Correct option A molecular mass of CaCO3 = (1 x 40) + (1 x 12) + (3 x 16) = 100 mass of CaCO3 given = 10 g no of moles in 10 g of CaCO3: 10 / 100 = 0.1 mol no of atoms in 0.1 mol = 6.02 x 1023 x 0.1 = 6.02 x 1022 Total no of protons in CaCO3: 20 (protons in Calcium) + 6 (protons in Carbon) + 3 x 8 (protons in oxygen) = 50 no of protons in 0.1 mol of CaCO3 = 6.02 x 1022 x 50 = 3.01 x 1024
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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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