You weigh out 0.1183 g of a complex salt to analyze 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 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).
H2O (l) + 2 MnO4- (aq) + 3 CN- (aq) ---> 2 MnO2 (s) + 3CNO- (aq) + 2 OH- (aq)
Pretty confused. A step-by-step answer, to help me understand, would be greatly appreciated. Thanks a ton.
H2O (l) + 2 MnO4- (aq) + 3 CN- (aq) ---> 2 MnO2 (s) + 3CNO- (aq) + 2 OH- (aq)
Pretty confused. A step-by-step answer, to help me understand, would be greatly appreciated. Thanks a ton.
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Well, I can take you a little farther, but the rest is up to you.
You have 10.02 mL of a 0.08035 mol/L solution of KMnO4- That is equal to:
10.02 x 0.08035 mol/L or 0.805107 millimoles.
Now the CN- reacts in a 3/2 mol ratio with KMnO4- so that ;means you will have used up
0.805107 x 3/2 = 1.2077 millimoles of CN-.
The molar mass of CN- is 26.06 g and 0.0012077 mol of that weighs 26.06 x 0.0012077 =0.0314 g
That's as far as I get. Hope it helps some.
You have 10.02 mL of a 0.08035 mol/L solution of KMnO4- That is equal to:
10.02 x 0.08035 mol/L or 0.805107 millimoles.
Now the CN- reacts in a 3/2 mol ratio with KMnO4- so that ;means you will have used up
0.805107 x 3/2 = 1.2077 millimoles of CN-.
The molar mass of CN- is 26.06 g and 0.0012077 mol of that weighs 26.06 x 0.0012077 =0.0314 g
That's as far as I get. Hope it helps some.