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Percent Yield with Limiting Reactant For the following reaction, 6.11 grams of s

ID: 503936 • Letter: P

Question

Percent Yield with Limiting Reactant For the following reaction, 6.11 grams of silver nitrate are mixed with 3.50 g Iron(II) chloride. The experiment actually yields 3.02 grams of iron(II) nitrate when performed in the lab. What is the theoretical yield of Iron(II) nitrate in grams? What is the percent yield of the experiment? iron(II) chloride (aq) + silver nitrate (aq) rightarrow iron(II) nitrate (aq) + silver chloride (s) Balanced equation _____ What is the theoretical yield of Iron(II) nitrate in grams? _____ What is the percent yield of the experiment? _____

Explanation / Answer

Solution:

The balanced reaction is as follows: FeCl2 (aq) + 2AgNO3 (aq) ------> Fe(NO3)2 (aq)+ 2 AgCl (s)

According to the stoichiometric balanced reaction 2 moles of AgNO3 reacts with 1 mol of FeCl2 to give 1 mol of Fe(NO3)2 as the product of the reaction.

Therefore molar ratio of the product, Fe(NO3)2 to the reactant, AgNO3 = 1/2

To calculate the theoretical yield of the product, Fe(NO3)2 we need to follow up the following equation.

grams product = grams reactant x (2 mol reactant/molar mass of reactant) x (mole ratio product/reactant) x (molar mass of product/1 mol product)

molar mass of AgNO3 = 169 g
molar mass of Fe(NO3)2 = 180 g

grams Fe(NO3)2 = grams AgNO3 x (2 mol AgNO3/169 g AgNO3) x (1 mol Fe(NO3)2 /2 mol AgNO3) x (180 g Fe(NO3)2 /1 mol Fe(NO3)2 )

We have 6.11 g of AgNO3, therefore

grams Fe(NO3)2 = 6.11 g AgNO3 x (2 mol AgNO3/169 g AgNO3) x (1 mol Fe(NO3)2 /2 mol AgNO3) x (180 g Fe(NO3)2 /1 mol Fe(NO3)2 )

grams Fe(NO3)2 = (6.11 x 2/169 x 1/2 x 180) g

grams Fe(NO3)2 = 6.5 g

Therefore, theoretical yield of Fe(NO3)2 is 6.5 g.

Percent yield of the experiment = Actual yield of the product x 100/Theoretical yield

                                                = 3.02x100/6.5 = 46.5%

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