If 5.65 grams of zinc metal react with 21.6 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 zn(no3)2 + ag

1.3710 grams of silver metal can be formed.

0.4158 grams of the excess reactant will be left over when the reaction is complete

Explanation:

Moles of zinc metal =

Moles of silver nitrate =

According to reaction , 1 mole of zinc reacts with 2 mole of silver nitrate.

Then 0.0864 mol of zinc will react with:

silver nitrate

This means that zinc is in excessive amount.

So, the formation of silver metal will depend on moles of silver nitrate.

According to reaction 2 moles of silver nitrate produces 2 mole of silver metal.

Then 0.01271 moles of silver nitrate will produce:

of silver metal

Mass of 0.01271 mol of silver metal = 0.01271 mol × 107.87 g/mol = 1.3710 g

Moles of excessive reagent left

According to reaction , 1 mole of zinc reacts with 2 mole of silver nitrate.

Then 0.01271 mol of silver nitrate will react with:

of zinc

Moles of an excess reagent = 0.01271 mol - 0.00635 mol = 0.00636 mol

Mass of 0.00636 mol zinc:

0.00636 mol × 65.38 g/mol = 0.4158 g

1.3710 grams of silver metal can be formed.

0.4158 grams of the excess reactant will be left over when the reaction is complete

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