Asked by Eman
If 200 ml of .300 molarity of silver nitrate are mixed with 350 ml of .500 molarity calcium chloride ions how many grams of percipitate are formed? How many moles of each ions are present?
Answers
Answered by
DrBob222
This is a limiting reagent problem. I know that because amounts for BOTH reactants are given.
2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
mols AgNO3 = M x L =?
mols CaCl2 = M x L = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols of the product.
Do the same for mols CaCl2 to mols of the product.
It is quite likely that the two answers will not agree; therefore, one of them is incorrect. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
I assume the problem is asking how many mols of each ion are present AFTER the reaction.
Ag and Cl must be determined by the Ksp for AgCl because AgCl is insoluble.
Ca and NO3^- are soluble.
2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
mols AgNO3 = M x L =?
mols CaCl2 = M x L = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols of the product.
Do the same for mols CaCl2 to mols of the product.
It is quite likely that the two answers will not agree; therefore, one of them is incorrect. The correct value in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
I assume the problem is asking how many mols of each ion are present AFTER the reaction.
Ag and Cl must be determined by the Ksp for AgCl because AgCl is insoluble.
Ca and NO3^- are soluble.
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