How many grams of each of the following compounds are present after the reaction is complete?

Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 8.00 g of sodium carbonate is mixed with one containing 6.75 g of silver nitrate. How many grams of each of the following compounds are present after the reaction is complete?

sodium carbonate

silver nitrate

silver carbonate

sodium nitrate

Answer 1

==Reaction Equation==
1 sodium carbonate+ 2 silver nitrates---> 1 silver carbonate+ 2 sodium nitrates

==Molecular mass==
sodium carbonate = 2*M(Na)+ M(C)+ 3*M(O) = 2*22.99+ 12.01+ 3*16.00 = 105.99 g/mol
silver nitrate = M(Ag)+ M(N)+ 3*M(O) = 107.9+ 14.01+ 3*16.00 =169.91 g/mol
silver carbonate = 2*M(Ag)+ M(C)+ 3*M(O) = 2*107.9+ 12.01+ 3*16.00 = 275.81 g/mol
sodium nitrate = M(Na)+ M(N)+ 3*M(O) = 22.99+ 14.01+ 3*16.00 = 85.00 g/mol

==Before reaction==
Amount in mols of sodium carbonate = 8.00g/(105.99g/mol) = 7.545*10^-2 mol [[01]]

Amount in mols of silver nitrate = 6.75/(169.91) = 3.973*10^-2 mol [[02]]

According to the reaction equation, for every unit of sodium carbonate, we need 2 of silver nitrate. In other words, we need twice as much silver nitrate as sodium carbonate.

However, comparing both [[01]] 7.545*10^-2 and [[02]] 3.973*10^-2, we figure out that we have pretty much about half as much silver nitrate as sodium carbonate. So silver nitrate is said to be the *limiting reactant*.

==Before reaction==
sodium carbonate........... = 7.545*10^-2 mol
silver nitrate..................... = 3.973*10^-2 mol = 2x <--- limiting reactant
silver carbonate............... = 0
sodium nitrate.................. = 0

==Variation due to reaction (according to the stoichiometric coefficients of the reaction equation)==
___ sodium carbonate........... = -x = -1.987*10^-2 mol
___ silver nitrate..................... = -2x = -3.973*10^-2 mol
___ silver carbonate............... = +x = +1.987*10^-2 mol
___ sodium nitrate.................. = +2x = +3.973*10^-2 mol

==After reaction (= before+ variation)==
sodium carbonate............ = 5.559*10^-2 mol----> 5.89 g
silver nitrate...................... = 0
silver carbonate................ = 1.987*10^-2 mol----> 5.48 g
sodium nitrate................... = 3.973*10^-2 mol----> 3.38 g