Pre-lab prep 4 com Complete Aeoiution containing 1.20g of Na:SO, was adod eold l

Question

Pre-lab prep 4 com Complete Aeoiution containing 1.20g of Na:SO, was adod eold lead (1) Pb(NO3)2. The products are sodium nitrate Answer the followin A solution containing 1.20g of .20g of Nasnitrate an .729- e mass of the solid collected after filtration and washing Collected ProdutsO, was added to another solution with 2.179 th o to another soltion with 2179 of tration m nitrate and solid lead (Il) sulfate. Th d washing was 0.572g a. Write the chemical equation equation for the reaction including state symbols. 14] b. Determine the limiting reagent. [4 c. What is the expected mass of lead (I) sulfate? [4]

Explanation / Answer

(a) The balanced molecular equation is:

Pb(NO3)2(aq) + Na2SO4(aq) -----> PbSO4(s) + 2NaNO3(aq)

(b) calculation of limiting reagent:

molar mass of Na2SO4 = 142.04g and mass = 1.2g

then moles of Na2SO4 = 1.2/142.04 = 0.008448 moloe

molar mass of Pb(NO3)2 = 331.21g and mass = 2.17g

then moles of  Pb(NO3)2 = 2.17/331.21 = 0.006552 moloe

Here moles of  Pb(NO3)2 ie less than moles of Na2SO4, hence  Pb(NO3)2 is the limiting reagent.

(c) expected mass of PbSO4:

As from balanced equation:

331.21g of  Pb(NO3)2 produces 303.26g of  PbSO4

then amount of  PbSO4 produced after reaction of 2.17 g of  Pb(NO3)2 = 2.17 x 303.26 / 331.21

= 658.07 / 331.21

= 1.9868g = 1.99g of PbSO4 (also theoretical yield of  PbSO4)

(d) percentage yield = experimental yield x 100 / theoretical yield

= 1.06 x 100 / 1.99

= 53.27%

(e)

(i) there may not be 100% precipitation of  PbSO4.

(ii) isolation of  PbSO4 may not be perfect (like filtration, drying etc).

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