Answer:
Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)
Explanation:
When lead (II) nitrate, Pb(NO3)2 undergoes a double displacement reaction with aqueous sodium chloride, NaCl, the following products are obtained:
Pb(NO3)2(aq) + NaCl(aq) —>
Aqueous Pb(NO3)2 will dissociate in solution as follow:
Pb(NO3)2(aq) —> Pb2+(aq) + 2NO3-(aq)
On the other hand, aqueous NaCl will dissociate as follow:
NaCl(aq) —> Na+(aq) + Cl-(aq)
The double displacement reaction will take place as follow:
Pb(NO3)2(aq) + NaCl(aq) —>
Pb2+(aq) + 2NO3-(aq) + Na+(aq) + Cl-(aq) —> Pb2+(aq) Cl-(aq) + Na+(aq) 2NO3-(aq)
Pb(NO3)2(aq) + NaCl(aq) —> PbCl2(s) + NaNO3(aq)
We simply balance the equation by putting 2 in front of NaCl and 2 in front of NaNO3 as shown below:
Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)
