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?
To determine the grams of precipitate formed, we need to identify the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed.
Step 1: Write the balanced chemical equation for the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2):
AgNO3 + CaCl2 → AgCl + Ca(NO3)2
Step 2: Calculate the moles of each reactant:
Moles of AgNO3 = Volume (L) × Molarity (mol/L)
Moles of AgNO3 = 0.200 L × 0.300 mol/L = 0.060 mol
Moles of CaCl2 = Volume (L) × Molarity (mol/L)
Moles of CaCl2 = 0.350 L × 0.500 mol/L = 0.175 mol
Step 3: Determine the limiting reactant by comparing the moles of each reactant. The reactant with the lower number of moles is the limiting reactant.
In this case, AgNO3 has 0.060 moles and CaCl2 has 0.175 moles. AgNO3 has fewer moles, so it is the limiting reactant.
Step 4: Use the stoichiometry of the balanced equation to determine the moles of AgCl formed.
From the balanced equation, we see that 1 mole of AgNO3 reacts to form 1 mole of AgCl. Therefore, 0.060 moles of AgNO3 will produce 0.060 moles of AgCl.
Step 5: Calculate the mass of AgCl formed using the molar mass of AgCl:
Molar mass of AgCl = 107.87g/mol
Mass of AgCl = Moles × Molar mass
Mass of AgCl = 0.060 mol × 107.87 g/mol = 6.47 g
Therefore, approximately 6.47 grams of precipitate (AgCl) are formed.
Now let's calculate the moles of each ion present:
Moles of Ag+ ions = 0.060 mol (since there is a 1:1 ratio of AgNO3 to Ag+ ions)
Moles of Cl- ions = 0.060 mol (since there is a 1:1 ratio of AgNO3 to Cl- ions)
Moles of Ca2+ ions = 0.175 mol (since there is a 1:1 ratio of CaCl2 to Ca2+ ions)
Moles of NO3- ions = 0.060 mol (since there is a 1:1 ratio of AgNO3 to NO3- ions)
So there are 0.060 moles of Ag+ ions, 0.060 moles of Cl- ions, 0.175 moles of Ca2+ ions, and 0.060 moles of NO3- ions present in the solution.
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.