Question

How many grams of each species will be present at the end when 0.500 g of copper is added to 2.500 g of AgNO3?

Answer 1

Answer : The mass of excess $AgNO_3$, $CuNO_3$  and Ag is, 1.16 g, 0.992 g and 0.849 g respectively.

Solution : Given,

Mass of Cu = 0.500 g

Mass of $AgNO_3$ = 2.500 g

Molar mass of Cu = 63.5 g/mole

Molar mass of Ag = 108 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $CuNO_3$ = 126 g/mole

First we have to calculate the moles of Cu and $AgNO_3$.

$\text{ Moles of }Cu=\frac{\text{ Mass of }Cu}{\text{ Molar mass of }Cu}=\frac{0.500g}{63.5g/mole}=0.00787moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{2.500g}{170g/mole}=0.0147moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Cu+AgNO_3\rightarrow CuNO_3+Ag$

From the balanced reaction we conclude that

As, 1 mole of $Cu$ react with 1 mole of $AgNO_3$

So, 0.00787 moles of $Cu$ react with 0.00787 moles of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and $Cu$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of excess $AgNO_3$

Moles of excess $AgNO_3$  = 0.0147 - 0.00787 = 0.00683 moles

Now we have to calculate the moles of $CuNO_3$  and Ag.

From the reaction, we conclude that

As, 1 mole of $Cu$ react to give 1 mole of $CuNO_3$

So, 0.00787 moles of $Cu$ react to give 0.00787 moles of $CuNO_3$

and,

As, 1 mole of $Cu$ react to give 1 mole of $Ag$

So, 0.00787 moles of $Cu$ react to give 0.00787 moles of $Ag$

Now we have to calculate the mass of excess $AgNO_3$, $CuNO_3$  and Ag.

$\text{ Mass of excess }AgNO_3=(0.00683moles)\times (170g/mole)=1.16g$

and,

$\text{ Mass of }CuNO_3=\text{ Moles of }CuNO_3\times \text{ Molar mass of }CuNO_3$

$\text{ Mass of }CuNO_3=(0.00787moles)\times (126g/mole)=0.992g$

and,

$\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag$

$\text{ Mass of }Ag=(0.00787moles)\times (108g/mole)=0.849g$

Thus, the mass of excess $AgNO_3$, $CuNO_3$  and Ag is, 1.16 g, 0.992 g and 0.849 g respectively.