Question

If 4.35 grams of zinc metal react with 35.8 grams of silver nitrate, how many grams of silver metal can be formed and how many grams of the excess reactant will be left over when the reaction is complete? Show all of your work.

unbalanced equation: Zn + AgNO3 “yields”/ Zn(NO3)2 + Ag

Answer 1

So far I have

Zn + 2 AgNO3 = Zn(NO3)2 + 2 Ag
4.35 g / 65.39 g/mol=.0665 mol Zn
.0665 * 2 = .133 mol AgNO3 required
35.8 g/ 169.87 g/mol=.2107 mol AgNO3
.2107 mol Ag
.2107 * 107.868 g/mol=22.7277876 g Ag
.2107/2 = .10535 mol Zn

Answer 2

Still need help?

Answer 3

yes, what do i do now?

Answer 4

Some of your steps were wrong.

Answer 5

ah, okay.

Answer 6

Remember, the numbers in front of the compounds are ratios of MOLES. Not mass. You should have divided the masses (4.35 and 65.39) by their respective molar masses.

Answer 7

I did that in my steps.

Answer 8

You divided 4.35 by 65.39. those are the masses of two different compounds.

Answer 9

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
Try this?

Answer 10

65.39 is the mass of Zinc.

Answer 11

169.87 is the mass of silver nitrate.

Answer 12

So what did I do wrong?

Answer 13

I checked the source. I still don't see what you mean.

Answer 14

@halorazer

Answer 15

Oh right. Duh. I'm an idiot. xD sorry. Let me try to work it out myself.

Answer 16

No, I make mistakes all the time. We're only human... Haha

Answer 17

http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.html
http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Limiting_Reagents
http://www.ausetute.com.au/exceslim.html
I don't wanna give you the wrong methods or whatever.

Answer 18

is it 35.8-22.7277876=