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How many ml of 0.117 M AgNO3 would be required to react exactly with the NaCl solution?
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∙ 9y ago
You need to know the amount of NaCl.
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∙ 8y ago
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How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?
Balanced equation first! AgNO3 + NaCl -> AgCl + NaNO3 all one to one, get moles AgNO3 3.82 moles NaCl (1 mole AgNO3/1 mole NaCl) = 3.82 moles AgNO3 ------------------------------- Molarity = moles of solute/Liters of solution 0.117 M AgNO3 = 3.82 moles AgNO3/Liters Liters = 3.82/0.117 = 32.6 Liters which is 32600 milliliters which is unreasonable; check answer if you can
Why carbon tetra chloride does not give white ppt with AgNO3 solution?
It may not react with silver nitrate, or if it does react, the product is soluble.
What type of reaction is AgNO3 plus NaOH?
They will form NaNO3 in aqueous solution, and AgOH would precipitate out of solution. AgNO3(aq) + NaOH(aq) --> AgOH(s) + NaNO3(aq) This is an example of a double displacement/replacement reaction.
What forms when you react KCl and AgNO3?
AgCl and KNO3.
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.
How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?
Balanced equation first! AgNO3 + NaCl -> AgCl + NaNO3 all one to one, get moles AgNO3 3.82 moles NaCl (1 mole AgNO3/1 mole NaCl) = 3.82 moles AgNO3 ------------------------------- Molarity = moles of solute/Liters of solution 0.117 M AgNO3 = 3.82 moles AgNO3/Liters Liters = 3.82/0.117 = 32.6 Liters which is 32600 milliliters which is unreasonable; check answer if you can
Why carbon tetra chloride does not give white ppt with AgNO3 solution?
It may not react with silver nitrate, or if it does react, the product is soluble.
What type of reaction is AgNO3 plus NaOH?
They will form NaNO3 in aqueous solution, and AgOH would precipitate out of solution. AgNO3(aq) + NaOH(aq) --> AgOH(s) + NaNO3(aq) This is an example of a double displacement/replacement reaction.
Can c2h2 react with agno3?
yes
What four ionic species existed before the solution were mixed for NaCl AgNO3?
Four ions exist: Na+, Cl-, Ag+, (NO3)-; sodium chloride react with silver nitrate:NaCl + AgNO3 = AgCl + NaNO3AgCl is insoluble in water.
What forms when you react KCl and AgNO3?
AgCl and KNO3.
What are the products when these react AlCl3 and HNO3 and AgNO3?
alcl3+hno3
Why is AgNo3 and NaCl least desirable?
These compounds react forming a white insoluble precipitate - silver chloride: NaCl + AgNO3 = AgCl + NaNO3
What volume of 0305 M AgNO3 is required to react exactly with 155.0ml of 0274 M Na2SO4 solution?
155.0 mL x 0.274 M SO42- reacts with 155.0 x 0.274 * 2 (Ag+/SO42-) = 84.94 mmol Ag+84.94 mmol Ag+ is in 84.94 (mmolAg+) / 0.305 (M Ag+) = 84.94/0.305 = 278.5 mL of 0.305 M Ag+.Na+ and NO3- are tribuned out of this reaction.
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 is the net ionic equation of AgNO3 HNO3?
Silver nitrate doesn't react with nitric acid.
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
