10.0 g of Zn powder is added into a solution of silver nitrate, AgNO3. The total mass of the metallic solid recovered at the end of the reaction is 12.32 g. Assuming that the reaction did not go to completion, how many grams of Zn did react?

 

So the equation I got what

Zn (s) + 2AgNO₃ (aq) -> 2Ag (s) + Zn(NO₃)₂ (s)

I think the zinc nitrate is solid because the question asks about a metallic solid.

 

Then,

12.32 g X 1 mol/189.41 g Zn(NO3)2 = .065 mol Zn(NO3)2

 

Then to find the original Zn,

.065 mol X 1 mol Zn/1mol Zn(NO3)2 = .065 mol Zn

 

.065 mol Zn X 65.39 g/1 mol = 4.25 g Zn

 

But the answer should be 1 g Zn. What did I do wrong?

Edited February 3, 201411 yr by rasen58

rasen58

 

Perhaps try googling zinc nitrate to see if one of your assumptions might not be correct. And what would happen to the silver?

 

But I am anxious for a chemist to stroll by because I cannot get 1 gram as an answer (I can get two grams - but I think that is wrong, see above comment)

12.32 grams of metal is sum of masses of zinc that didn't react plus silver that was made from silver nitrate.

12.32 grams of metal is sum of masses of zinc that didn't react plus silver that was made from silver nitrate.

 

Yep - that makes sense and now my sums work out.