Potassium hexachloroferrate(II) can be dissolved in water and reacted with cyani

Question

Potassium hexachloroferrate(II) can be dissolved in water and reacted with cyanide ion (CN-) to form the more favorable cyano-complex forms: K4[FeCl6] + 6 CN- ===> K4[Fe(CN)6] + 6 Cl- A student performs this experiment by adding 20 mL of 4.0 M NaCN (aq) to 4.562 g of K4[FeCl6]. The solid product was isolated, dried, and massed. Answer the following questions based on this information.
1. How many moles of potassium hexachloroferrate(II) are available at the start?
2. How many moles of sodium cyanide are added?
3. Which reactant is the limiting reagant?
4. What is the theoretical yield of K4[Fe(CN)6] in grams?
5. The mass of the completely dried K4[Fe(CN)6] was 4.213 g. What was this student's % yield?
Potassium hexachloroferrate(II) can be dissolved in water and reacted with cyanide ion (CN-) to form the more favorable cyano-complex forms: K4[FeCl6] + 6 CN- ===> K4[Fe(CN)6] + 6 Cl- A student performs this experiment by adding 20 mL of 4.0 M NaCN (aq) to 4.562 g of K4[FeCl6]. The solid product was isolated, dried, and massed. Answer the following questions based on this information.
1. How many moles of potassium hexachloroferrate(II) are available at the start?
2. How many moles of sodium cyanide are added?
3. Which reactant is the limiting reagant?
4. What is the theoretical yield of K4[Fe(CN)6] in grams?
5. The mass of the completely dried K4[Fe(CN)6] was 4.213 g. What was this student's % yield?
K4[FeCl6] + 6 CN- ===> K4[Fe(CN)6] + 6 Cl- A student performs this experiment by adding 20 mL of 4.0 M NaCN (aq) to 4.562 g of K4[FeCl6]. The solid product was isolated, dried, and massed. Answer the following questions based on this information.
1. How many moles of potassium hexachloroferrate(II) are available at the start?
2. How many moles of sodium cyanide are added?
3. Which reactant is the limiting reagant?
4. What is the theoretical yield of K4[Fe(CN)6] in grams?
5. The mass of the completely dried K4[Fe(CN)6] was 4.213 g. What was this student's % yield?

Explanation / Answer

Number of moles of K4[Fe(Cl)6] = 4.562 / 424.956 = 0.01073 moles
number of moles of NaCN = 20 * 4/1000 = 0.08 moles
K4[Fe(Cl)6] is limiting reagent since it has less number of moles.
1 mole of K4[Fe(Cl)6 produces 1 mole of K4[Fe(CN)6
number of moles of K4[Fe(CN)6] = 0.01073 moles
mass of K4[Fe(CN)6] = 0.01073 * 368.35 = 3.952 g

percentage yield = observed mass/ theoritical mass *100

                                = 4.213/3.952 * 100 = 106.6%

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