silver nitrate reacts with iron(III) choride to give silver chloride and iron (I

Question

silver nitrate reacts with iron(III) choride to give silver chloride and iron (III) nitrate.in a particular experiment it was planned to mic a solution containing 35.0 g of AgNO3 with another solution containing 50.0 g of FeC13. a. Write the chemical equation for the reaction. b.Which reactant is the limiting reactant? c. What is the maximum number of moles of AgC1 that could be obtained from this mixture? d.What is the maximum number of grams of AgC1 that could be obtained? e.How many grams of reactant in excess will remain after the reaction is over? f. I the yield of the reaction is 28.50 g, calculate the % yield of this reaction.

Explanation / Answer

A) Chemical Equation=   3AgNO3+ FeCl3 ---> 3AgCl + Fe(NO3)3

b) To find the limiting reactant:

35.0g AgNO3 x (1 mol AgNO3 / 169.87g AgNO3) x (3 mol AgCl / 3 mol AgNO3) x (143.32g AgCl/ 1 mol AgCl)

= 29.53g AgCl

Next use the 50.0g of FeCl3:

50.0g FeCl3 x (1 mol FeCl3/ 162.2g FeCl3) x (3 mol AgCl / 1 mol FeCl3) x (143.32g AgCl / 1 mol AgCl)

= 132.54g AgCl

B) Therefore, AgNO3 is the limiting reactant, as it would run out faster in the reaction.

c) The maximum number of moles of AgCl that could be produced in this reaction =

35.0g AgNO3 x (1 mol AgNO3 / 169.87g AgNO3) x (3 mol AgCl / 3 mol AgNO3) = 0.206 moles AgCl

C) Therefore, the maximum number of moles that could be produced in this reaction = 0.206 moles AgCl

D) The maximum number of grams of AgCl that could be obtained (calculated previously) = 29.53g AgCl

e)

Determine moles of limiting reactant: 35g AgNO3 x (1 mol AgNO3 / 169.87g AgNO3) = 0.206 moles AgNO3

Then, use molar ratios to determine how many grams of excess (FeCl3) will remain:

0.206 moles AgNO3 x (1 mol FeCl3 / 3 mol AgNO3) x (162.2g/ 1 mol FeCl) = 11.14 g FeCl3

Then subtract the amount of excess calculated above from the given amount of FeCl3 to find the amount remaining:

50.0g FeCl3 - 11.14g FeCl3 = 38.86 g FeCl3

E) The amount of Excess (FeCL3) that will remain in excess when the reaction is over = 38.86g FeCl3

f)

Percent yield = (Actual yield / Theoretical yield) (x100)

The theoretical mass that was calculated for AgCL using the limiting reactatant = 29.53g AgCl

Therefore, (28.50g AgCl / 29.53g AgCl) x 100 = 96.5 %

F) The Percent yield for this equation = 96.5 %

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