2) The following series of reactions were carried out.Pb(NO3)2(aq) + H2O() + CO2(g)PbCO3(s) + 2HNO3 (aq)Pb(NO3)2(aq) + 2HCI(aq)> 2HNO3(aq) + PbCl2(s)other reagents are added in excess, what is the theoretical yield of lead(II) chloridesolid?(a) If a student starts with 3.000 g of lead(II) carbonate for the first reaction and allIMOIPDC03-O3.000gPhto2 2012 1Peco0.12o Po CO 3.0112motPoCat ImoiPolNo.O112mol PbNO22.Imol Pbcos2HNO2Lo12molPolNo3) maPbN)0224 mol O0224mol HNO3 2.003a02I HNO31.3924NO3(b) If the student isolates 2.571g of lead(II) chloride, what is the percent yield?2.51lgl1,342

Question
Asked Oct 4, 2019
2) The following series of reactions were carried out.
Pb(NO3)2(aq) + H2O() + CO2(g)
PbCO3(s) + 2HNO3 (aq)
Pb(NO3)2(aq) + 2HCI(aq)> 2HNO3(aq) + PbCl2(s)
other reagents are added in excess, what is the theoretical yield of lead(II) chloride
solid?
(a) If a student starts with 3.000 g of lead(II) carbonate for the first reaction and all
IMOIPDC03-O
3.000gPhto2 2012 1Peco0.12o Po CO 3
.0112motPoCat ImoiPolNo
.O112mol PbNO2
2.
Imol Pbcos
2HNO2
Lo12molPolNo3) maPbN)0224 mol O
0224mol HNO3 2.003a02
I HNO3
1.3924NO3
(b) If the student isolates 2.571g of lead(II) chloride, what is the percent yield?
2.51lgl
1,342
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2) The following series of reactions were carried out. Pb(NO3)2(aq) + H2O() + CO2(g) PbCO3(s) + 2HNO3 (aq) Pb(NO3)2(aq) + 2HCI(aq)> 2HNO3(aq) + PbCl2(s) other reagents are added in excess, what is the theoretical yield of lead(II) chloride solid? (a) If a student starts with 3.000 g of lead(II) carbonate for the first reaction and all IMOIPDC03-O 3.000gPhto2 2012 1Peco0.12o Po CO 3 .0112motPoCat ImoiPolNo .O112mol PbNO2 2. Imol Pbcos 2HNO2 Lo12molPolNo3) maPbN)0224 mol O 0224mol HNO3 2.003a02 I HNO3 1.3924NO3 (b) If the student isolates 2.571g of lead(II) chloride, what is the percent yield? 2.51lgl 1,342

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Step 1

(a)

 

Given that the mass of PbCO3 is 3.0 g. From the balanced equation, it can be seen that number of moles of lead carbonate will be equal to the number of moles of lead chloride.

PbCO
2HNO> Pb(NO;), + H2O+CO2
Pb(NO) 2HCi-2HNO+PbCl
No.of moles of PbCO,-molar mass
mass
3g
11
267.21
=0.0112mol
No.of moles of PbCO3= No.of moles of PbCl, -0.0112 mol
_
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PbCO 2HNO> Pb(NO;), + H2O+CO2 Pb(NO) 2HCi-2HNO+PbCl No.of moles of PbCO,-molar mass mass 3g 11 267.21 =0.0112mol No.of moles of PbCO3= No.of moles of PbCl, -0.0112 mol _

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Step 2

Mass of PbCl2 can be calculated...

Mass of PbCl, No.of molesxmolarmass
0.0112molx278.19g/mol
-3.12g
Theoretical yield of PbCl2=3.12g
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Mass of PbCl, No.of molesxmolarmass 0.0112molx278.19g/mol -3.12g Theoretical yield of PbCl2=3.12g

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