Question #8e3d3

1 Answer
anor277
Aug 3, 2017

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..

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