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ipn [44]
3 years ago
5

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)
Chemistry
1 answer:
Stels [109]3 years ago
8 0

Answer:

25.35%

Explanation:

Again let me restate the the equation of the reaction;

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

Amount of potassium permanganate  reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles

If 2 moles of MnO4 - reacts with 3 moles of CN-

8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2

= 1.229 * 10^-3 moles of CN-

Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol

= 0.03 g

Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100

= 25.35%

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Type the correct answer in the box. Express your answer to three significant figures.
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<u>Given:</u>

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

<u>To determine:</u>

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<u>Explanation:</u>

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

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Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g



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Answer:

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