Question:
What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl ? (Atomic mass Ca = 40, Ag = 108)
What mass of calcium chloride in grams would be enough to produce 14.35 g of AgCl ? (Atomic mass Ca = 40, Ag = 108)
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Always ensure the chemical equation is balanced to get the correct mole ratio between the reactants and products.
Updated On: May 2, 2026
- 5.55 g
- 8.295 g
- 11.19 g
- 16.59 g
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
This is a stoichiometry problem based on the precipitation reaction between calcium chloride and silver nitrate.
Step 2: Detailed Explanation:
The balanced chemical equation is: \[ CaCl_2 + 2AgNO_3 \rightarrow 2AgCl + Ca(NO_3)_2 \] Molar mass of \(AgCl\) = \(108 + 35.5 = 143.5\) g/mol. Moles of \(AgCl\) formed = \(\frac{14.35}{143.5} = 0.1\) mol. From the equation, 2 moles of \(AgCl\) are produced from 1 mole of \(CaCl_2\). So, moles of \(CaCl_2\) required = \(\frac{0.1}{2} = 0.05\) mol. Molar mass of \(CaCl_2\) = \(40 + (2 \times 35.5) = 40 + 71 = 111\) g/mol. Mass of \(CaCl_2\) required = \(0.05 \times 111 = 5.55\) g.
Step 3: Final Answer:
The required mass of calcium chloride is 5.55 g, option (A).
This is a stoichiometry problem based on the precipitation reaction between calcium chloride and silver nitrate.
Step 2: Detailed Explanation:
The balanced chemical equation is: \[ CaCl_2 + 2AgNO_3 \rightarrow 2AgCl + Ca(NO_3)_2 \] Molar mass of \(AgCl\) = \(108 + 35.5 = 143.5\) g/mol. Moles of \(AgCl\) formed = \(\frac{14.35}{143.5} = 0.1\) mol. From the equation, 2 moles of \(AgCl\) are produced from 1 mole of \(CaCl_2\). So, moles of \(CaCl_2\) required = \(\frac{0.1}{2} = 0.05\) mol. Molar mass of \(CaCl_2\) = \(40 + (2 \times 35.5) = 40 + 71 = 111\) g/mol. Mass of \(CaCl_2\) required = \(0.05 \times 111 = 5.55\) g.
Step 3: Final Answer:
The required mass of calcium chloride is 5.55 g, option (A).
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