Asked by Jefferson
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.
Answers
Answered by
Devron
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.
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.
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