Stoichiometry Problems for Reaction in Aqueous Solutions -. The quantity of Cl –

Question

Stoichiometry Problems for Reaction in Aqueous Solutions

-.The quantity of Cl– in a municipal water supply is determined by titrating the sample with Ag+. The precipitation reaction taking place during the titration is Ag(aq) + Cl(aq) ->AgCl(s)

The end point in this type of titration is marked by a change in color of a special type of indicator. (a) How many grams of chloride ion are in a sample of the water if 20.2 mL of 0.200 M Ag+ is needed to react with all the chloride in the sample? (b) If the sample has a mass of 15.0 g, what percentage of Cl– does it contain?

-.A mysterious white powder is found at a crime scene. A simple chemical analysis concludes that the powder is a mixture of sugar and morphine (C17H19NO3), a weak base similar to ammonia. The crime lab takes 10.00 g of the mysterious white powder, dissolves it in 100.00 mL water, and titrates it to the equivalence point with 2.84 mL of a standard 0.100 M HCl solution. What is the percentage of morphine in the white powder?

-.A sample of an iron ore is dissolved in acid, and the iron is converted to Fe2+. The sample is then titrated with 57.20 mL of 0.05240 M MnO4– solution. The oxidation-reduction reaction that occurs during titration is

MnO4(aq) + Fe^2+(aq) +8H(aq) -> Mn^2+(aq) +5Fe^3+(aq) + 4H2O(l)

How many moles of MnO4– were added to the solution? (b) How many moles of Fe2+ were in the sample? (c) How many grams of iron were in the sample? (d) If the sample had a mass of 2.8890 g, what is the percentage of iron in the sample?

-A sample of 70.5 mg of potassium phosphate is added to 25.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) Calculate the theoretical yield, in grams, of the precipitate that forms.

Explanation / Answer

A.

0.0202 Lx0.200 mol/L = 0.00404 mol Ag+

and the same quantity is for Cl-.

a)0.00404 mol Cl- x 35.45 g/mol = 0.143 g Cl-

b)100x0.143 g /15 g = 0.956 %

B.

0.00284 L x 0.1 mol/L = 0.000284 mol morphinr in the sample.

Molar mass of morphine 285.34 g/mol

0.000284 mol x 285.34 g/mol = 0.081 g

100x0.081 g/10.0 g = 0.81 % morphine in sample

C.

a. 57.20 mL x 0.05240 M MnO4– = 3 mmol MnO4–

Correct your equation:

MnO4-(aq) + 5Fe^2+(aq) +8H(aq) -> Mn^2+(aq) +5Fe^3+(aq) + 4H2O(l)

b. The molar ratio Fe/Mn is 5/1, so

3 mmol x 5 = 15 mmol Fe

c. 0.015 mol x 55.85 = 0.83775 g Fe

d. 100 x0.83775 g/ 2.8890 g = 29.0 %

D.

a. 3AgNO3 + K3PO4 = Ag3PO4 +3 KNO3

b. Molar mass K3PO4 136.086 g/mol

0.0705 g/ 136.086 g/mol = 0.000518 mol K3PO4

0.025 L x 0.050 mol/L = 0.00125 mol AgNO3

0.00125 mol AgNO3 is < 3x 0.000518

AgNO3 is the limiting reagent

c.

3AgNO3 + K3PO4 = Ag3PO4 +3 KNO3

3 mol………………….1 mol

0.00125 mol…………..x

X = 0.000417 mol Ag3PO4

418.5760 g/mol x 0.000417 mol = 0.174 g Ag3PO4

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