200cm\(^3\) each of 0.1M solutions of lead (II) trioxonitrate(V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated.
[Pb = 207, Cl= 35.5, N = 14, O = 16]
2.78g
5.56g
8.34g
11.12g
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equation is pb( no3)2+2hcl->pbcl2 +2hno3
no of mole= conc.*volume/1000
no of mole=0.1*200/1000
=0.02mols
from moles=mass/molar mass
0.02=m/278
m=5.56g
Pb(NO3)2 (aq) + 2HCL (aq) - pbcl2 (s) + 2HNO3 (aq)
Molar mass of pbcl2=207+2*35.5=278gmol
1000cm3 contains 0.1M
1cm3 contains 0.1M/1000
200cm3 contains 0.1*200/1000=5.56g
2HCl + Pb(NO3)2 ===> 2HNO3 + PbCl2
2moles of HCl ===> 1mole of PbCl2
n = CV/1000 = (100 x 0.2)/(1000) = 0.02 mol
n (HCl) = 0.02 mol
2 moles of HCl ===> 1 mole of PbCl2
0.02 mole of HCl ===> x mole of PbCl2
2x = 0.02
x = 0.01 mol
n = m/M
m = 0.01 x (207 + (35.5 x 2)) = 0.01 x 278 = 2.78 g
dnt get confused h2so4+nacl----hcl+naoh+o2 so you will get a huge, corrosive, white fumes of HCl gas being produced.
Sori d formula to gt mass concentrn is molar concentrn * molar mass while dt of molar concntrn is mass concentrn/molar mass so pls correct frm wat I said earlier.
m/mm = cv/1000
m=?
m.m of PbCl2=207+2(35.5)
= 207+71= 278
C= 0.1
V=200
m/278 = 0.1×200/1000
m/278 = 20/1000
CROSS MULTIPLY ✖️
=1000×m = 278× 20
=1000 m = 5560
DBS by 1000
m= 5.56(ANS)
The balanced equation is: Pb(NO3)2 + 2HCl = PbCl2 + 2HNO3
using : m= CVM , m is the mass in g=? , C is molar concentration = 0.1M, M is molar mass of required compound = 207 + 71 = 278g/mol and V is volume in dm3= 0.2dm3
m= 0.1 x 0.2 x 278= 5.56g
based on the balanced equation one mole of HCl produced one mole of PbCl2 not 2 mole produced one mole. pls go through the solution again there are some mistakes
