Question

In: Chemistry

Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride solution until...

Silver sulfate solution is added to 25.00 mL of a 0.500 M potassium chloride solution until no more precipitate forms. What mass of silver chloride will be formed? What concentration of potassium will remain in solution after the reaction is complete if 45 mL of silver sulfate solution were added to the potassium chloride solution?

Solutions

Expert Solution

2KCl(aq) + Ag2SO4(aq)----> 2AgCl(s) + 2KNO3(aq)
2 mole                   2 mole

no of moles of KCl = Molarity * volume in ml
                    = 0.5 * 0.025
                    = 0.0125 moles of KCl
from balanced equation
2 mole of KCl react with Ag2So4 to from 2 mole of AgCl
0.0125 mole of KCl react with AgNO3 to from 0.0125 moles of AgCl
mass of AgCl = no of moles of AgCl * molar mass of AgCl
             = 0.0125*143.4 = 1.793 gm of AgCl

KCl
molarity M1   = 0.5 M
volume V1     = 25ml
no of mole n1 = 2

Ag2SO4
molarity M2    =
volume    V2   = 45ml
no of moles n2 = 1

M1V1/n1    =   M2V2/n2

M2          = M1V1n2/n1V2
            = 0.5*25*1/2*45
            = 0.138M
concentration of KCl left = 0.5-0.138 = 0.368 M


Related Solutions

When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate,...
When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction occurs. Write the balanced net ionic equation of the reaction.
A 17.0 mL sample of a 1.62 M potassium sulfate solution is mixed with 14.8 mL...
A 17.0 mL sample of a 1.62 M potassium sulfate solution is mixed with 14.8 mL of a 0.890 M barium nitrate solution and this precipitation reaction occurs: K2SO4(aq)+Ba(NO3)2(aq)→BaSO4(s)+2KNO3(aq) The solid BaSO4 is collected, dried, and found to have a mass of 2.48 g . Determine the limiting reactant, the theoretical yield, and the percent yield. Part A Determine the limiting reactant. Express your answer as a chemical formula. K2SO4+Ba(NO3)2→BaSO4+2KNO3 SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining Part...
A 65.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 34.0 mL...
A 65.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 34.0 mL of a 0.100 M lead (II) acetate solution and the following precipitation reaction occurs: K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 0.994 g . Determine the limiting reactant, the theoretical yield, and the percent yield.
A 62.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 36.5 mL...
A 62.0 mL sample of a 0.122 M potassium sulfate solution is mixed with 36.5 mL of a 0.108 M lead(II) acetate solution and the following precipitation reaction occurs: K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 0.992 g . Determine the limiting reactant, the theoretical yield, and the percent yield.
A 29.3-mL sample of a 1.84 M potassium chloride solution is mixed with 15.0 mL of...
A 29.3-mL sample of a 1.84 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2KNO3(aq) What is the 1. Limiting reactant 2. Theoretical yield 3. percent yield This seems like an easy question but for some reason I keep getting hung up on these!
Aqueous potassium phosphate is added to 545 mL of a solution containing calcium chloride to precipitate...
Aqueous potassium phosphate is added to 545 mL of a solution containing calcium chloride to precipitate all of the Ca2+ ions as the insoluble phosphate (310.2 g/mol). If 4.51 g calcium phosphate is produced, what is the Ca2+ concentration in the calcium chloride solution in g/L?
if a titration using these solutions started with 25.00 mL of potassium chromate solution (0.5056 M),...
if a titration using these solutions started with 25.00 mL of potassium chromate solution (0.5056 M), what volume of 0.2456 M lead (II) nitrate solution would be required to reach the end point? what MASS of the products would be created in that reaction?
Calculate the expected change in temperature for 25.00 mL of 0.500 M HCl reacting with 25.00...
Calculate the expected change in temperature for 25.00 mL of 0.500 M HCl reacting with 25.00 mL of 0.500 M NaOH. The heat Capacity of the Calorimeter is 15.6 J C -1 and the molar Enthalpy of Neutralization is -55.83 kJ mole . (Remember to use -1 dimensional analysis)
The molar solubility of silver sulfate in a 0.142 M ammonium sulfate solution is _______ M
The molar solubility of silver sulfate in a 0.142 M ammonium sulfate solution is _______ M  
Solid sodium iodide is slowly added to 175 mL of a 0.468 M silver acetate solution until the concentration of iodide ion is 0.0687 M.
Solid sodium iodide is slowly added to 175 mL of a 0.468 M silver acetate solution until the concentration of iodide ion is 0.0687 M. The mass of silver ion remaining in solution is _______  gramsSolid chromium(III) acetate is slowly added to 75.0 mL of a 0.388 M potassium hydroxide solution until the concentration of chromium(IIID ion is 0.0556 M. The percent of hydroxide ion remaining in solution is _______ % 
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT