?55.0mL sample of a 0.110 M potassium sutato souton is moed wth 380mL of a 0.124

Question

?55.0mL sample of a 0.110 M potassium sutato souton is moed wth 380mL of a 0.124 Mlead m acetate sch tion and re folomg pred tuton-don ooor The solild Phsca is collected, dried, and found to have a mass of 1.01 Part A identi?y the limiting reactant K2504 P504 Previous Answers Correct The imiting reactant of a reacton is found by calouilsting the number of moles of each reactant and comparing them The reactant with the least number of moies is considered the nving reactant because onceallf the reaction stope despite the presence of another reactant. The inting reactant is imporlant when delermining tre heoretical yleid of a produst Part Determine the theoretical yiald mass of P804 Reqvest Anawer Part C Determine the percent yield.

Explanation / Answer

Number of moles of K2SO4 = Volume of solution (in L ) * Molarity(M) = 55/1000 * 0.110 = 0.00605 moles

Number of moles of Lead (II) acetate = Volume of solution (in L ) * Molarity(M) = 38/1000 * 0.24 = 0.004712 moles

Hence limiting reagent is 0.004712 moles i.e. Lead (II) acetate

One mole of lead acetate gives one mole of lead sulphate

Hence number of lead sulphate formed = 0.004712 moles

Molar mass of Lead sulphate = 303.26 gm/mol

Theoritical Yield = Number of moles * Molar mass

=> 0.004712 mol * 303.26 g/mol = 1.4289 grams

Part C

Percent Yield = Actual Yield/Theoritical Yield * 100

=> 1.01/1.4289 * 100

=> 70.68*

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