[onetwothree123]

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 the reaction did not go to completion, how many grams of Zn did react?

What exactly constitutes the 12.32g and how do you get the final answer of 1.00?

[mikasaur]

Perhaps start by writing out the chemical equation. What are the reactants and products here?

[onetwothree123]

The reactants should be Zn and AgNO₃ and the products should be ZnNO₃ and Ag. I know i have to use an X variable to denote one of the products but I'm not too sure how..

[mikasaur]

Good start, but double check your products. What is the chemical formula for the compound containing zinc and nitrate?

Once you figure that out, remember that atoms cannot be created or lost in a reaction. It all balances out in the end... 

[onetwothree123]

It should be Zn(NO₃)₂ . which means there will be 2 Ag in the products and 2 silver nitrate in reactant. So does that mean 12.32 represents the Zn in the Zinc nitrate + the silver?

[mikasaur]

The last step that people often miss in a balanced chemical equation are the states of each entity. The question was nice enough to tell you that you start with zinc powder and a solution of silver nitrate. If your possible states are solid, liquid, gas, aqueous... what would the states for the two reactants be? How about your products?

If you're concerned with the mass of the metallic solid recovered, what must that be?

[onetwothree123]

Zinc powder is a solid. Silver nitrate is aqueous. Silver is a solid. Zinc Nitrate is always soluble. However the reaction did not go to completion so is it a solid actially? The mass of the metallic solid should be theunreacted zinc and silver at the end

[Arkcon]

Zinc powder is a solid. Silver nitrate is aqueous. Silver is a solid. Zinc Nitrate is always soluble. However the reaction did not go to completion so is it a solid actially? The mass of the metallic solid should be theunreacted zinc and silver at the end

Nope.  Almost, but not quite.  Please draw out the reaction in chemical shorthand, with phases.  This is an important first step, and you've skipped it three times already, and just typed words.  Once you write it out in chemical shorthand, you will likely never again claim that silver is present.  Or at least, you'll call it a solution of silver nitrate in water, which is not a solid.

[onetwothree123]

Oh I see the net ionic without the nitrate would end up like Zn(s) + 2Ag(s) -> 2Ag(s) + Zn(s) since not all the silver goes to completion...
so if x were the zinc not used then 10-x would be zinc used, my equation would be
x + (10-x)/MM Zinc · 2 mol Ag/1 mol Zn · MM Ag = 12.37