Answer to Question #73724 in General Chemistry for aleia

Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 5.50 g of sodium carbonate is mixed with one containing 2.50 g of silver nitrate. After the reaction is complete, the solutions are evaporated to dryness, leaving a mixture of salts. How many grams of each of the following compounds are present after the reaction is complete?


sodium carbonate

g

silver nitrate

g

silver carbonate

g

sodium nitrate

g

Answer 1

m(Na2CO3)=5.5 g
.m(AgNO3)=2.5 g
Find m (Na2CO3)=?
.m(AgNO3)=?
.m(Ag2CO3)=?
.m(NaNO3)=?
The reaction is: Na2CO3 +2 AgNO3=Ag2CO3+ 2NaNO3
Mr(Na2CO3)= 2 Ar(Na) +Ar(C) +3 Ar(O)=2*23+12+3*16=106
Mr(AgNO3)=Ar(Ag) +Ar(N)+ 3Ar(O)=108+14+3*16=170
Mr(Ag2CO3) = 2 Ar(Ag)+Ar(C) +3Ar(O)=2*108+12+3*16=276
Mr(NaNO3)=Ar(Na)+Ar(N)+3Ar(O)=23+14+3*16=85
Find amount of substance n(Na2CO3)=m (Na2CO3)/Mr(Na2CO3)=5.5/106=0.052 mol
.n(AgNO3)=m(AgNO3)/Mr(AgNO3)=2.5/170=0.015 mol
Na2CO3 is the excess reactant n (Na2CO3)=n(NaCO3)-1/2 n(AgNO3)=0.052-0.0075=0.0445 mol
.m(Na2CO3)=n*Mr=0.0445*106=4.72 g
AgNO3 is the limiting reactant
.n(Ag2CO3)=1/2n(AgNO3)=1/2*0.015=0.0075 mol
.m(Ag2CO3)=n(Ag2CO3)*Mr(Ag2CO3)=0.0075*276=2.07 g
.n(NaNO3)=n(AgNO3)=0.015 mol
.m(NaNO3)=n(NaNO3) Mr(NaNO3)=0.015*85=1.275 g