Question #8e3d3

1 Answer
Aug 3, 2017

Answer:

#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..