±Exercise 4.70 with feedback A 60.0 mL sample of a 0.114 M potassium sulfate sol

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±Exercise 4.70 with feedback

A 60.0 mL sample of a 0.114 M potassium sulfate solution is mixed with 38.0 mL of a 0.128 M lead(II) acetate solution and the following precipitation reaction occurs:

K2SO4(aq)+Pb(C2H3O2)2(aq)2KC2H3O2(aq)+PbSO4(s)

The solid PbSO4 is collected, dried, and found to have a mass of 1.00 g .

Determine the limiting reactant, the theoretical yield, and the percent yield.

Part A

Identify the limiting reactant.

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The limiting reactant of a reaction is found by calculating the number of moles of each reactant and comparing them. The reactant with the least number of moles is considered the limiting reactant because once all of it has reacted the reaction stops despite the presence of another reactant. The limiting reactant is important when determining the theoretical yield of a product.

Part B

Determine the theoretical yield.

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Part C

Determine the percent yield.

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General Chemistry - Chem 50

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Chapter Four Homework

±Exercise 4.70 with feedback

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Explanation / Answer

Solution :-

Lets first calculate the moles of the both reactants

Moles = molarity * volume in liter

Moles of K2SO4 = 0.114 mol per L * 0.060 L

                          = 0.00684 mol

Moles of Pb(C2H3O2)2 = 0.128 mol per L * 0.0380 L

                                     = 0.004864 mol

Mole ratio of the both reactant is 1 :1

Therefore the reactant with lower mole is the limiting reactant

Therefore Pb(C2H3O2)2 is the limiting reactant.

Now lets calculate the theoretical yield of the product PbSO4 using the mole ratio of the limiting reactant and product

Mole ratio of the Pb(C2H3O2)2 and PbSO4 is 1 :1 therefore the moles of the PbSO4 produced are same as moles of Pb(C2H3O2)2

Therefore moles of PbSO4 = 0.004864 mol

Now lets convert these moles to mass

Mass = moles * molar mass

Mass of PbSO4 = 0.004864 mol * 303.262 g per mol

                            = 1.475 g PbSO4

Therefore the theoretical yield = 1.475 g

Now lets calculate the percent yield

% yield = (actual mass / theoretical mass)*100%

              = (1.00 g / 1.475 g)*100%

              = 67.80 %

Therefore percent yield = 67.80 %

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