Question

Question #8e3d3

Answer 1

`Zn(s) + 2AgNO_3(aq) rarr Zn(NO_3)_2(aq) + 2Ag(s)darr`

Explanation:

We have `(2.00*g)/(65.4*g*mol^-1)=0.0306*mol` `"zinc metal"`.

And `(2.50*g)/(169.87*g*mol^-1)=0.0147*mol` `"silver nitrate"`.

And so the zinc metal reductant is slightly in stoichiometric excess.

We form `2xx0.0147*molxx107.9*g*mol^-1=3.17*g` with respect to silver metal.

I will let you answer the last question. There are `1.2*mmol` of EXCESS zinc metal..