Name Total SHOW ALL WORK NO WORK = NO CREDIT e l. You weigh out 0.1183 g of a co

Question

Name Total SHOW ALL WORK NO WORK = NO CREDIT e l. You weigh out 0.1183 g of a complex salt to analyze for the percentage of cyanide ion in your complex salt. pfe ai ertfo A fter dissolving the complex salt in solution, you determine that it takes 10.02 mL of a 0.08035 M potassium permanganate solution to react with all of the cyanide ion in your complex salt. Using the balanced chemical equation provided, determine the mass percentage of the cyanide ion in your complex salt. (Remember: the mole ratio of MnO4 to KMnO4 is 1:1) H20 (0) 2 MnO, (aq) 3 CN (aq) 2 MnO2 (s) 3 CNO (aq) 2 OH (aq)

Explanation / Answer

Hello,

The following questions pertain to Stoichiometric Calculations & can be solved in the following manner:

1) You weigh out, 0.1183 g of complex salt to analyse for the percentage of cyanide ion in your complex salt. After dissolving the complex salt in solution, you determine that it takes 10.02 ml of a 0.08035 M potassium permanganate solution to react with all the cyanide ion in your complex salt. Using the balanced chemical equation provided, determine the mass percentage of the cyanide ion in your complex salt.

H2O(l) + 2MnO4-(aq) + 3CN-(aq) = 2MnO2 (s) + 3CNO- + 2OH-(aq)

This question can be solved using two methods:

a) The equivalents method:

Pros: You don’t need to balance the reaction

Con: Slightly difficult

To use this method, you’ll have to remember that for any reaction the number of equivalents of all the combining species is the same as the number of equivalents of all the products formed.

Also, number of equivalents = weight given/equivalent weight

& equivalent weight = molecular weight/ n-factor

For an oxidation or reduction reaction the n-factor is equal to the number of moles of electrons lost or gained respectively, per mole of reactant.

In this reaction:

a) MnO4- has Mn in +7 oxidation state, which undergoes reduction to MnO2 which has Mn in +4 oxidation state. So, for this change, 3 moles of electrons are lost for every mole of MnO4- during its conversion to MnO2, then n-factor is equal to 3.

b) CN- has carbon in +2 oxidation state, which undergoes reduction to CNO- which has carbon in +4 oxidation state. So, for this change, 2 moles of electrons are lost for every mole of CN- during its conversion to CNO-, then n-factor is equal to 2.

Next,

Millimoles of MnO4- = Molarity x Volume = 0.08035 M x 10.02 ml = 0.8051 millimoles

Milliequivalents of MnO4- = Millimoles x n-factor = 0.8051 x 3 = 2.415 milliequivalents

Milliequivalents of CN- = 2.415 milliequivalents

Weight of CN- in the sample = Equivalents of CN- x Equivalent weight of CN- = 2.415 x 10-3 x 13 g = 0.0313 g [equivalent weight of CN- = Molar weight of CN-/n-factor = 26/2 = 13]

Therefore, mass percentage of CN- in the sample = (0.0313 g/0.1183 g) x 100 = 26.46 %

b) The mole method:

Pros: Easy method

Con: You’ll have to properly balance the reaction to use this method

To use this method you’ll have to remember that if a balanced reaction is represented as:

aA + bB = cC + dD

Then we can write: (moles of A/a) = (moles of B/b) = (moles of C/c) = (moles of D/d)

So, for the reaction between permanganate ion and cyanide ion, we can write:

Moles of MnO4-/2 = Moles of CN-/3

(Since the ratio of moles of MnO4- & CN- is 2/3)

Millimoles of MnO4- = molarity x volume = 0.08035 M x 10.02 ml = 0.8051 millimoles

So, moles of CN? = [(0.8051 x 3)/2] x 10-3 = 0.001207 moles

Weight of CN- = 0.001207 moles x 26 g = 0.0313 g

Therefore, mass percentage of CN- in the sample = (0.0313 g/0.1183 g) x 100 = 26.46 %

2) You want to determine the amount of Fe (III) in a complex salt. You make a 100 ml solution of the complex salt by dissolving 0.0670 g of complex salt with water to the mark on a 100 ml volumetric flask. You get an absorbance reading for the iron from the spectrometer of 0.334. The molar absorptivity is determined to be 17605 L/mole.cm and the path length is 1 cm. Determine the molarity of iron(III) in the solution. Determine the mass percentage of iron(III) in the complex salt.

To solve this question, we need to use the Beer’s law. Beer’s law governs the amount of radiation absorbed and suggests that the absorbance is proportional to the concentration. As the concentration of the dissolved compound increases in the solvent, the absorbance increases proportionately. So, according to this law:

Absorbance = Molar absorptivity x Path Length x Concentration

Using this equation, we’d obtain:

Molar Concentration = [Absorbance/(path length x molar absorptivity)] = 0.334/(17605 L/mole.cm x 1 cm) = 1.89 x 10-5 moles.L-1

1000 ml of solution would thus have 1.89 x 10-5 moles of Fe(III)

100 ml of solution would have 0.189 x 10-5 moles of Fe(III)

Weight of Fe(III) derived from complex salt = 0.189 x 10-5 moles x atomic weight of Fe = 0.189 x 10-5 moles x 56 g = 0.000106 g

Therefore, mass percentage of Fe(III) in the complex salt is = (0.000106 g/0.0670 g) x 100 = 0.158 %

Hope this helps. I request you to take some time tp rate the answer & to drop in your valuable feedback.
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