Find the limiting and excess reagent

Question

Silver is expensive. One way to recover silver from waste silver chloride produced in an investigation is to react it with zinc metal in a basic solution:

2 AgCl(s) + 4 NaOH(aq) + Zn(s) → Na2Zn(OH)4(aq) + 2 Ag(s) + 2 NaCl(aq)

Choose the limiting and excess reagents in this process so that all of the silver is extracted from the silver chloride. Justify your choices.

Answer 1

To get all of the Ag out of the AgCl, AgCl needs to be the limiting reactant. Then, NaOH and Zn are both the excess reactants.

2 AgCl(s) + 4 NaOH(aq) + Zn(s) → Na₂Zn(OH)₄(aq) + 2 Ag(s) + 2 NaCl(aq)

The best way to do this is to consider moles of reaction. Moles of reaction is how many times the reaction can occur, basically, and is given by:

moles reaction = ^{moles of substance present}/ _{coefficient of substance}

And is given by the coefficients, so that:

1 mol reaction = 2 mol AgCl = 4 mol NaOH = 1 mol Zn

The limiting reactant is that which corresponds to the fewest moles of reaction. So, just make AgCl correspond to 1 mol of reaction, and make the NaOH and Zn correspond to more than 1 mol of reaction. So, let's make AgCl be 1 mol reaction, and make NaOH and Zn correspond to 2 moles of reaction, and figure out those amounts:

1 mol reaction * (2 mol AgCl / 1 mol reaction) = 2 mol AgCl

2 mol reaction * (4 mol NaOH / 1 mol reaction) = 8 mol NaOH

2 mol reaction * (1 mol Zn / 1 mol reaction) = 2 mol Zn

Therefore, when the reaction is done, there will be 0 moles of AgCl remaining, and ALL of the Ag in the AgCl will exist only as Ag(s). Here is how the moles of each substance will change. In the ICE table,

I=initial,

C=change, which depends on coefficients,

and

E=end

...2 AgCl(s) + 4 NaOH(aq) + Zn(s) → Na₂Zn(OH)₄(aq) + 2 Ag(s) + 2 NaCl(aq)

I....2mol..........8mol.............2mol..........0mol.....................0mol...........0mol

C......-2x..........-4x...............-x...............+x.........................+2x............+2x

E....2-2x........8-4x...........2-x..............0+x...........................2x...............2x

Since AgCl is the limiting reactant, at the End, there are 0 moles of AgCl remaining. So, for AgCl:

2-2x=0

Solve for x.

x=1

Therefore, for the C row, x=1

...2 AgCl(s) + 4 NaOH(aq) + Zn(s) → Na₂Zn(OH)₄(aq) + 2 Ag(s) + 2 NaCl(aq)

I....2mol..........8mol.............2mol..........0mol.....................0mol...........0mol

C......-2(1)..........-4(1)...............-1..............+1........................+2(1)............+2(1)

E....0mol...........4mol............1mol...........1mol.....................2mol............2mol

So, if AgCl is limiting, then all of the Ag from AgCl ends up as Ag(s). There will be excess NaOH and Zn at the end.