Respuesta :
The balanced chemical reaction:
Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess
Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3
The limiting reactant is AgNO3.
0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2
0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess
0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess
Answer : The mass of excess reactant will be left over is 4.03 g and mass of [tex]Cu(NO_3)_2[/tex] is 34.2 g.
Solution : Given,
Mass of Cu = 9.85 g
Mass of [tex]AgNO_3[/tex] = 31.0 g
Molar mass of Cu = 64 g/mole
Molar mass of [tex]AgNO_3[/tex] = 170 g/mole
Molar mass of [tex]Cu(NO_3)_2[/tex] = 188 g/mole
First we have to calculate the moles of Cu and [tex]AgNO_3[/tex].
[tex]\text{ Moles of }Cu=\frac{\text{ Mass of }Cu}{\text{ Molar mass of }Cu}=\frac{9.85g}{64g/mole}=0.154moles[/tex]
[tex]\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{31.0g}{170g/mole}=0.182moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]AgNO_3[/tex] react with 1 mole of [tex]Cu[/tex]
So, 0.182 moles of [tex]AgNO_3[/tex] react with [tex]\frac{0.182}{2}=0.091[/tex] moles of [tex]Cu[/tex]
From this we conclude that, [tex]Cu[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.
Moles of remaining excess reactant = 0.154 - 0.091 = 0.063 moles
Now we have to calculate the mass of excess reactant will be left over.
[tex]\text{ Mass of }Cu=\text{ Moles of }Cu\times \text{ Molar mass of }Cu[/tex]
[tex]\text{ Mass of }Cu=(0.063moles)\times (64g/mole)=4.03g[/tex]
Now we have to calculate the moles of [tex]Cu(NO_3)_2[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]AgNO_3[/tex] react to give 1 mole of [tex]Cu(NO_3)_2[/tex]
So, 0.182 moles of [tex]AgNO_3[/tex] react to give [tex]\frac{0.182}{2}=0.091[/tex] moles of [tex]Cu(NO_3)_2[/tex]
Now we have to calculate the mass of [tex]Cu(NO_3)_2[/tex]
[tex]\text{ Mass of }Cu(NO_3)_2=\text{ Moles of }Cu(NO_3)_2\times \text{ Molar mass of }Cu(NO_3)_2[/tex]
[tex]\text{ Mass of }Cu(NO_3)_2=(0.182moles)\times (188g/mole)=34.2g[/tex]
Therefore, the mass of [tex]Cu(NO_3)_2[/tex] is 34.2 grams.
