A strip of zinc medal weighing 2.00 grams is placed in an aqueous solution containing 150.ml of .750M Tin(II) nitrate. Determine the grams of tin metal that will be precipitated.

Answer 1

The reaction is as followed:

Zn + Sn(NO3)2 --> Sn + Zn(NO3)2

Convert 0.750M Sn(NO3)2 to moles:

Molarity (M)= moles/volume (L)

Molarity*Volume (L)=moles

moles of Sn(NO3)2=0.750M*0.150L

This is similar to the limiting reagent problem. Proceed as you would do for the limiting reagent problem.