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3 years ago

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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 g

rams 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

2 answers:

Igoryamba3 years ago

7 0

The balanced chemical reaction is:

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

To determine the amount of the reactant left, we have to determine which is the limiting and the excess reactant. We do as follows:

5.65 g Zn ( 1 mol / 65.38 g) = 0.09 mol Zn
21.6 g AgNO3 (1 mol / 169.87 g) = 0.13 mol AgNO3

The limiting reactant would be silver nitrate since it is consumed completely in the reaction. The excess reactant would be zinc.

Excess Zinc = 0.09 mol Zn - (0.13 / 2) mol Zn = 0.025 mol Zn left

STALIN [3.7K]3 years ago

5 0

Answer:

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:

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

Moles of zinc metal = $\frac{5.65 g}{65.38 g/mol}=0.0864 mol$

Moles of silver nitrate = $\frac{21.6}{169.87 g/mol}=0.01271 mol$

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

Then 0.0864 mol of zinc will react with:

$2\times 0.0864 mol=0.1728 mol$ 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:

$\frac{2}{2}\times 0.01271 mol=0.01271 mol$ 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:

$\frac{1}{2}\times 0.01271 mol=0.00635 mol$ 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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