&.o/7.14 points | Previous Answers ZumChemP7 4.E 039 My Notes Ask Your How many

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&.o/7.14 points | Previous Answers ZumChemP7 4.E 039 My Notes Ask Your How many grams of silver chloride can be prepared by the reaction of 108.5 mL of 0.22 M silver nitrate with 108.5 mL of 0.14 M calcium chloride? 3.153 Calculate the concentrations of each ion remaining in solution after precipitation is complete. Ag 0239 NOj Ca2+ Need Help?ReadWatch Supporting Materials Periodic Table Constants and Factors Supplemental Data 9. 7.14 points ZumChemP7 4.E.041 My Notes Ask Your What mass of Na2CrO4 is required to precipitate all of the silver ions from 74.4 mL of a 0.120 M solution of AgNO3? Need Help? Read Wstch Supporting Materials Periodic Table Constants and Factors

Explanation / Answer

First lets write the reaction for part A

Silver nitrate is AgNO3

calcium chloride CaCl2

2 AgNO3 + CaCl2 ===== 2 AgCl + Ca(NO3)2

calculate the moles for each specie

108.5 ml or 0.1085 liters

Molarity of AgNO3 0.22 M

moles of AgNO3 = Molarity * volume = 0.1085 * 0.22 = 0.02387 moles

for CaCl2

0.1085 L and 0.14 M

moles of CaCl2 = 0.1085 * 0.14 = 0.01519 moles

The reaction is 2 to 1 , we need 2 moles of AgNO3 for every mole of CaCl2 so

there are 0.02387 moles of AgCl so we need the half of this of CaCl2 so

0.02387 / 2 = 0.0119 moles, there are 0.01519 moles of CaCl2 so there is an excess, the AgNO3 is the limiting reactant

since 2 moles of AgNO3 produces 2 moles of AgCl so there will be 0.02387 moles of AgCl

molar mass of AgCl is 143.32 g/gmol

mass = moles * molar mass 0.02387 * 143.32 =3.42 grams of AgCl formed

Ag reacts to form AgCl since it was the limiting reactant then there is no Ag in solution

Ag in solution is zero

Initial moles of Cl = 2 * 0.01519 = 0.03038 moles , 2 moles for every CaCl2

moles of Cl being precipitated = 0.02387 moles of Cl precipitated

Cl = 0.03038 - 0.02387 = 0.00651 moles of Cl in solution

Concentration is 0.00651 / 0.217 = 0.03 M, the 217 ml is the sum of the volumess 108.5 + 108.5

The nitrate is soluble in water so all the initial moles of nitrate will be present so

moles of NO3 = 0.02387 moles

[NO3] = 0.02387 / 0.217 = 0.11 M

Calcium is the same case

moles of calcium

0.01519 / 0.217 = 0.07 M

These last 2 do not precipitate they remain in solution

B)

The reaction is

Na2CrO4 + 2 AgNO3 ====== Ag2CrO4 + 2 NaNO3

calculate the number of moles with

moles = molarity * volume

moles = 0.12 * 0.0744 = 0.008928

according to the reaction we need 1 mole for every 2 moles of AgNO3 so we need the half amount of the AgNO3 available so

0.008928 / 2 = 0.04464 moles

mass = moles * molar mass

mass = 0.04464 * 162 = 0.72 grams of Na2CrO4 needed

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