Question #105006
How many grams of lead nitrate are needed to produce 105.25 g of sodium nitrate?
1
Expert's answer
2020-03-09T09:45:14-0400

Pb(NO3)2+2NaOH=2NaNO3+Pb(OH)2Pb(NO_3)_2 + 2NaOH = 2NaNO_3 + Pb(OH) _2

ν(NaNO3)=m/M=105.25/85=1.24mol.\nu(NaNO_3) = m/M = 105.25/85 = 1.24 mol.

1.24 : 2 = ν(Pb(NO3)2)\nu(Pb(NO_3)_2) : 1

ν(Pb(NO3)2)=1.24/2=0.62mol.\nu(Pb(NO_3)_2)=1.24/2=0.62 mol. m(Pb(NO3)2)=ν∗M=0.62∗331=205.22g.m(Pb(NO_3)_2)=\nu*M=0.62*331=205.22 g.


Answer: 205.22 g.


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