# How can zinc metal still be present after the reaction? [closed]

A $$\pu{15.8 g}$$ strip of zinc metal was placed in $$\pu{100 ml}$$ of silver nitrate. When the reaction was complete, the strip of zinc had a mass of $$\pu{13.1 g}$$. What was the concentration of the silver nitrate solution?

I am confused on writing the equation for this problem, because doesn't single displacement occur? So, how can zinc metal still be there?

## closed as off-topic by Klaus-Dieter Warzecha, Philipp, user2117, tschoppi, MichielMar 23 '14 at 15:49

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• There is an excess of zinc, so some is left over. Look at the problem again with that in mind. – Nicolau Saker Neto Dec 2 '13 at 2:14
• This is a flawed problem. In these types of redox displacements, the silver will plate onto the zinc strip. – Ben Norris Mar 23 '14 at 11:12

As we can clearly see, the silver nitrate is the limiting reactant in this problem as there are still 13.1 grams of zinc left that have not reacted.

That means that $$\pu{15.8 g}$$ $$\ce{Zn}$$ metal - $$\pu{13.1 g}$$ $$\ce{Zn}$$ metal = $$\pu{2.7 g}$$ $$\ce{Zn}$$ metal has reacted with all of the $$\ce{100 mL}$$ silver nitrate solution with $$\pu{13.1 g}$$ excess zinc metal.

We are however given $$\pu{100 mL}$$ of silver nitrate, and we are interested to find out how much product is formed. Looking online, the density of silver nitrate is $$\pu{5.35 g/cm3}$$ and keeping in mind that $$\pu{1 cm3} = \pu{1 mL}$$. Note that since the silver nitrate is in solution, the density of the silver nitrate in solution is unknown, but for references sake, we will refer to the density of the solid silver nitrate which is the 5.35 grams per cubic centimeter.

Knowing that $$\rho = m/V,$$ we can find the mass by $$\pu{5.35 g/cm^3} = \frac{m}{\pu{100 mL} \times \pu{1 cm^3/mL}},$$ therefore our mass of silver nitrate (theoretically) is $$m = \pu{535 g}.$$

In looking at the reactivity series, zinc is higher than silver, so single displacement does occur just as you predicted and therefore zinc "should not" appear as it then performs a redox reaction with the nitrate in silver nitrate.

See Wikipedia for the reactivity series of the metals.

If we write the chemical equation for this reaction we can see that $$\ce{AgNO3(aq) + Zn(s) -> ZnNO3(aq) + Ag(s)},$$ and the balanced equation is: $$\ce{2AgNO3(aq) + Zn(s) -> Zn(NO3)2(aq) + Ag(s)}.$$

Let's quantitatively calculate what would happen if we reacted each of our reactants.

Assuming that zinc is the limiting reactant and we have silver nitrate in excess:

$$\pu{15.8 g}\ \ce{Zn} \times \frac{\pu{1 mol}\ \ce{Zn}}{\pu{65.38 g}\ \ce{Zn}} \times \frac{\pu{1 mol}\ \ce{Zn}}{\pu{1 mol}\ \ce{Ag}} \times \frac{\pu{1 mol}\ \ce{Ag}}{\pu{107.9 g}\ \ce{Ag}} = \pu{0.002240 g}\ \ce{Ag}$$

Assuming that silver nitrate is the limiting reactant and we have zinc in excess: $$\pu{535 g}\ \ce{AgNO3} \times \frac{\pu{1 mol}\ \ce{AgNO3}}{\pu{169.9 g}\ \ce{AgNO3}} \times \frac{\pu{1 mol}\ \ce{Ag}}{\pu{2 mol}\ \ce{AgNO3}} \times \frac{\pu{107.9 g}\ \ce{Ag}}{\pu{1 mol}\ \ce{Ag}} = \pu{169.9 g}\ \ce{Ag}$$

In both cases, we expect to see aqueous zinc nitrate to form, and solid silver to precipitate. In looking at my example, all of the zinc is consumed in the reaction so we expect to not see any zinc metal left in the first reaction since that is the limiting reactant, because there is not enough zinc metal to react with the silver nitrate. In the second reaction, we see that silver nitrate is in excess because we have a much greater yield for silver, however, this reaction is "limited" due to not having enough zinc metal.

However, in your case, 13.1 grams of zinc metal are still remaining. Therefore, zinc must be the reactant in excess and thus the mass of silver nitrate is all consumed in the reaction. Therefore although you expected the zinc metal to bind to the nitrate in the redox reaction, it does! However, we have too much zinc metal and not enough silver nitrate, therefore we still see some zinc metal left along with our usual products of zinc nitrate and silver. I just did a basic stoichiometric limiting reactant problem so you can quantitatively see how this type of reaction works. In reality, I do not know the density of the silver nitrate solution, and thus the mass of the silver nitrate, which is much lower in the context of your problem.

• To me it doesn't make any sense to find the mass of 100 mL of solid silver nitrate. This answer is overly complicated for a problem, which is somewhat simple... – Martin - マーチン Oct 8 '18 at 10:03