roblem 3.82-Enhanced-with Feedback Part A Constants 1 Periodic Table Solutions o
ID: 1089024 • Letter: R
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roblem 3.82-Enhanced-with Feedback Part A Constants 1 Periodic Table Solutions of suiltfuric acid and leadill) acetate react to form solid leadll) suifate and a solution of acetic acid. 5.70 g of sulfuric acid and 5.70 g of lead) acetate are mixed. Calculate the number of grams of sulfuric acid present in the miature after the reaction is complete. Express your answer using three significant figures You may want to reference (Pages 106 -110) Section 3.7 while completing this problem. Submit Part B Calculate the number of grams of lead]I) acetate present in the mixture after the reaction is complete Express your answer using three significant figures. Part C Caloulabe the number of grams of lead)l) sutlate prosent in the mixture ater the reacton is complete. Express your answer using three significant figuresExplanation / Answer
H2SO4(aq) + (CH3COO)2Pb(aq) ---------> PbSO4(s) + 2CH3COOH(aq)
no of moles of H2SO4 = W/G.M.Wt
= 5.7/98 = 0.058 moles
no of moles of (CH3COO)2Pb = W/G.M.Wt
= 5.7/325 = 0.0175 moles
part-B
1 mole of H2So4 react with 1 mole of (CH3COO)2Pb
0.058 moles of H2SO4 react with 0.058 moles of (CH3COO)2Pb
(CH3COO)2Pb is limiting reactant
The no of moles of (CH3COO)2Pb reamains after comlete the reaction = 0.0175-0.0175 = 0
the mass of (CH3COO)2Pb remains after complete the reaction = 0
part-A
1 mole of (CH3COO)Pb react with 1 mole of H2So4
0.0175 moles of (CH3COO)2Pb react with 0.0175 moles of H2SO4
H2SO4 is excess reactant
The no of moles of excess after complete the reaction = 0.058-0.0175 = 0.0405 moles of H2So4
The amount of exces after complete the reaction = no of moles * gram molar mass
= 0.0405*325 = 13.1625g of H2SO4
Part-C
1 mole of (CH3COO)2Pb react with H2So4 to gives 1 mole of PbSo4
0.0175 moles of (CH3COO)2Pb react with H2SO4 to gives 0.0175 moles of PbSO4
mass of PbSO4 = no of moles * gram molar mass
= 0.0175*303 = 5.3g of PbSO4
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