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A solution containing 14 grams of AgNo3 is added to a solution containing 4.83 g of CaCl2. Find the mass of the precipitate produced.?
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2
Precipitate = AgCl
Now find the mol of compound in each solution:
14g AgNO3 x (mol/170g) = .082mol
4.83g CaCl2 x (mol/111g) = .044mol
Determine limiting reactant:
Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant.
Now that you know this you can find the mass of the precipitate
.082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g
b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left
.006mol x (111g/mol) = 0.67g CaCl2
What else can I help you with?
What is the test reagent for chloride ion?
Silver nitrate (AgNO3) is commonly used as the test reagent for chloride ions. When silver nitrate is added to a solution containing chloride ions, a white precipitate of silver chloride (AgCl) forms. This reaction is often used to detect the presence of chloride ions in a solution.
What volume of a 0.20 M silver nitrate solution is required to precipitate all the Cl - ion in the solution as AgCl?
To precipitate all the chloride ions in solution, you would need a 1:1 ratio of Ag+ to Cl-. The molar ratio between silver nitrate (AgNO3) and chloride ions (Cl-) is 1:1. So, you would need the same volume of 0.20 M AgNO3 solution as the volume of the chloride ions solution to precipitate all the Cl-.
When solutions NACl and AgNO 3 are mixed what precipitate forms?
NaCl(aq) + AgNO3(aq) = AgCl(s) + NaNO3(aq) - so the precipitate is white silver chloride.
What happen when aqueous bromide and aqueous silver nitrate mixed?
When aqueous bromide and aqueous silver nitrate are mixed, a white precipitate of silver bromide is formed due to a double displacement reaction. The balanced chemical equation for this reaction is: AgNO3(aq) + KBr(aq) → AgBr(s) + KNO3(aq)
What precipitates out in this reaction of sodium chromate and silver nitrate?
In this reaction, the precipitate formed would be silver chromate due to the double displacement reaction between sodium chromate (Na2CrO4) and silver nitrate (AgNO3). Silver chromate is insoluble in water, so it will precipitate out of the solution as a solid, appearing as a yellow precipitate.
What test reagent will distinguish a solube Cl- salt from So4-2?
A solution of a soluble chloride will give a white precipitate (turning purple on exposure to light) with silver nitrate solution. Sulfates do not react. Alternatively, the solution of sulfate will give a white precipitate with barium chloride solution, and the chloride solution will not.
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
What is the reagent used to precipitate chloride ions?
Silver nitrate (AgNO3) is commonly used to precipitate chloride ions as silver chloride (AgCl) in a chemical reaction. When a solution containing chloride ions is mixed with silver nitrate, a white precipitate of silver chloride forms.
Will Nibr2 and AgNO3 form a precipitate?
Yes, Nibr2 and AgNO3 will form a precipitate when mixed. This reaction is a double displacement reaction where the insoluble silver bromide (AgBr) precipitate will form in solution.
How many mol of AgCl can be precipitated by adding a solution containing 0.100 mol of AgNO3 to a solution containg excess NaCl?
One mole of AgNO3 reacts with one mole of NaCl to form one mole of AgCl precipitate. Therefore, 0.100 mol of AgNO3 will form 0.100 mol of AgCl precipitate when reacted with excess NaCl.
Why agno3 titration?
AgNO3 titration is commonly used to determine the concentration of chloride ions in a solution. Silver nitrate (AgNO3) reacts with chloride ions to form a white precipitate of silver chloride. The amount of AgNO3 required to completely precipitate all the chloride ions can be used to calculate the concentration of chloride in the solution.
How can you make potassium chloride precipitate?
You can make potassium chloride precipitate by adding silver nitrate (AgNO3). The chemical equation being AgNO3(aq)+ KCl(aq) = KNO3(aq) + AgCl(s) You know that silver nitrate will form a precipitate as you can see this on a solubility chart.
Will a precipitate form if solutions of the soluble salts AgNO3 and NaOH are mixed?
Yes, a white precipitate of silver hydroxide (AgOH) will form when solutions of silver nitrate (AgNO3) and sodium hydroxide (NaOH) are mixed. Silver hydroxide is insoluble in water, so it will precipitate out of the solution.
Will AgNO3 and NaCl form a precipitate when mixed?
Yes, when AgNO3 and NaCl are mixed, they will form a white precipitate of silver chloride (AgCl) because silver chloride is insoluble in water. This reaction is often used to confirm the presence of chloride ions in a solution.
Would AgNO3 and NH3 make a muddy brown precipitate?
When aqueous AgNO3 and NH3 are mixed, they react to form a white precipitate of silver(I) oxide (Ag2O), not a muddy brown precipitate. Silver oxide is insoluble in water and appears as a white solid. Any muddy appearance could be due to impurities or other reactions occurring in the solution.
What will happen when 2 drops of the NaCl solution is mixed with 2 drops of the AgNO3 solution?
A white precipitate of AgCl will form as a result of the reaction between sodium chloride (NaCl) and silver nitrate (AgNO3). The reaction can be represented by the equation: NaCl + AgNO3 → AgCl + NaNO3.
How do you test for presence of iodide ion?
You can test for the presence of iodide ions using silver nitrate. When silver nitrate is added to a solution containing iodide ions, a yellow precipitate of silver iodide forms. This precipitate confirms the presence of iodide ions in the solution.
