Asked by Nicole
Based on the following equation: 2AgNO3 + CaCl2 > 2AgCl + Ca(NO3) When 50 g AgNO3 are combined with 1.45M CaCl2 solution, what volume of the 1.45M CaCl2 solution would be needed to precipitate all of the silver as AgCl?
Answers
Answered by
DrBob222
mols AgNO3 = grams/molar mass = approx 0.3 but you need a better answer than that estimate.
Convert mols AgNO3 to mols CaCl2. That means approx 0.3 x (1 mol CaCl2/2 mols AgNO3) = approx 0.15 mol CaCl2.
Then M CaCl2 = mols CaCl2/L CaCl2. You have M CaCl2 and mols CaCl2, solve for L CaCl2. Covert to mL if needed.
Convert mols AgNO3 to mols CaCl2. That means approx 0.3 x (1 mol CaCl2/2 mols AgNO3) = approx 0.15 mol CaCl2.
Then M CaCl2 = mols CaCl2/L CaCl2. You have M CaCl2 and mols CaCl2, solve for L CaCl2. Covert to mL if needed.
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