Respuesta :
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
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
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:
[tex]Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag[/tex]
Moles of zinc metal =[tex]\frac{5.65 g}{65.38 g/mol}=0.0864 mol[/tex]
Moles of silver nitrate = [tex]\frac{21.6}{169.87 g/mol}=0.01271 mol[/tex]
According to reaction , 1 mole of zinc reacts with 2 mole of silver nitrate.
Then 0.0864 mol of zinc will react with:
[tex]2\times 0.0864 mol=0.1728 mol[/tex] 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:
[tex]\frac{2}{2}\times 0.01271 mol=0.01271 mol[/tex] 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:
[tex]\frac{1}{2}\times 0.01271 mol=0.00635 mol[/tex] 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
