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

The balanced chemical reaction is written as:

Zn + 2AgNO3 = Zn (NO3) 2 + 2Ag

To determine the grams of silver metal that is being produced, it is important to first determine which is the limiting reactant and the excess reactant from the given initial amounts. We do as follows:

4.35 g Zn (1 mol / 65.38 g) (2 mol AgNO3 / 1 mol Zn) = 0.1331 mol AgNO3 needed

35.8 g AgNO3 (1 mol / 169.87 g) (1 mol Zn / 2 mol AgNO3) = 0.1054 mol Zn needed

Therefore, the limiting reactant would be the zinc metal since it would be consumed completely in the reaction. The excess amount of AgNO3 would be:

0.2107 mol AgNO3 - 0.1331 mol AgNO3 = 0.0776 mol AgNO3 left (169.87 g / 1 mol) = 13.19 g AgNO3 left

0.0665 mol Zn (2 mol Ag / 1 mol Zn) (107.9 g / 1 mol) = 14.3581 g Ag produced