Multiple choice

A sample of mixture of $CaCl_{2}$ and $NaCl$ weighing $5.55$ g was treated to precipitate all the $Ca$ as $CaCO_{3}$ which was then heated and quantitatively converted to $1.68$ g of $CaO$. The mass percentage of $CaCl_{2}$ in the mixture is;

  1. $60$%
  2. $70$%
  3. $45$%
  4. $56$%
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A Correct answer
Explanation

1.68g of CaO corresponds to 1.68 / 56 = 0.03 moles of Ca. This Ca came from CaCl2. Moles of CaCl2 = 0.03. Mass of CaCl2 = 0.03 * 111 = 3.33g. Percentage = (3.33 / 5.55) * 100 = 60%.

AI explanation

Calculate the moles of CaO precipitated by dividing 1.68 g by its molar mass of 56 g/mol, yielding 0.03 moles. Because one mole of CaCl2 produces one mole of CaO, the original mixture contains 0.03 moles of CaCl2. Multiplying this by the molar mass of CaCl2 (111 g/mol) gives a mass of 3.33 g, and dividing this by the total 5.55 g mixture mass gives a mass percentage of 60%.

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