les an e Top t 14 Synthesis Homework 4/2912017 11:55 PM S7.2/10 4/18/2017 o3:49
ID: 517735 • Letter: L
Question
les an e Top t 14 Synthesis Homework 4/2912017 11:55 PM S7.2/10 4/18/2017 o3:49 PM Grade Print BCalaulator Periodic Table Sapling Learning this question has been astomiradby Donna McGregora Cty University of New York KUNm Lehman An aqueous solution containing 9.98 g of Iead(l) nitrate is added to an containing 7.19 g of aqueous solution potassium chloride to generate solid lead(ll)chloride and potassium nitrate. the balanced chemical equation for this reaction. Be sure to include all physical states. Tip: you need to clear your work and reset the equation, click the button that looks like two arrows. 'What is the limiting reactant? O lead(II) nitrate tium chloride The percent yield for the reaction is 89.3%, how many grams of precipitate were recovered? 7A8 How many grams of the excess reactant remain? A Previous Give Up & View soluton Try Again 0 Nama Exit O Explanation areers MacBook AirExplanation / Answer
Pb(NO3)2(aq) + 2KCl(aq) --------------> PbCl2(s) + 2KNO3(aq)
no of moles of Pb(NO3)2 = W/G.M.Wt = 9.96/331.2 = 0.03 moles
no of moles of KCl = W/G.M.Wt = 7.19/74.5 = 0.0965 moles
1 mole of Pb(NO3)2 react with 2 moles of KCl
0.03 moles of Pb(NO3)2 react with = 2*0.03/1 = 0.06 moles of KCl
Pb(NO3)2 is limiting reactant
1 mole of Pb(NO3)2 react with KCl to gives 1mole of PbCl2
0.03 moles of Pb(NO3)2 react with KCl to gives = 1*0.03/1 = 0.03 moles of PbCl2
mass of PbCl2 = 0.03*278 = 8.34g
percentage yield = actual yield*100/theoritical yield
89.3 = x*100/8.34
X = 89.3*8.34/100 = 7.48 g
excess reactant is KCl
No of moles of excess reaqctant remains = 0.0965-0.06 = 0.0365 moles
mass of excess reactant remains = 0.0365*74.5 = 2.72g
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