Question

When 40.0 g of copper are reacted with silver nitrate solution Cu + 2 AgNO3 --> Cu(NO3)2 + 2 Ag 118 g of silver are obtained. What is the percent yield of silver? molar mass of silver = 107.9 g, molar mass of copper = 63.55 g

Answer 1

Answer: The percent yield of silver is 86.96 %

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of copper = 40.0 g

Molar mass of copper = 63.55 g/mol

Putting values in equation 1, we get:

`\text{Moles of copper}=(40.0g)/(63.55g/mol)=0.629mol`

The given chemical equation follows:

`Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag`

By Stoichiometry of the reaction:

1 mole of copper produces 2 moles of silver

So, 0.629 moles of copper will produce =
`(2)/(1)* 0.629=1.258mol` of silver

Now, calculating the mass of silver from equation 1, we get:

Molar mass of silver = 107.9 g/mol

Moles of silver = 1.258 moles

Putting values in equation 1, we get:

`1.258mol=\frac{\text{Mass of silver}}{107.9g/mol}\\\\\text{Mass of silver}=(1.285mol* 107.9g/mol)=135.7g`

To calculate the percentage yield of silver, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of silver = 118 g

Theoretical yield of silver = 135.7 g

Putting values in above equation, we get:

`\%\text{ yield of silver}=(118g)/(135.7g)* 100\\\\\% \text{yield of silver}=86.96\%`

Hence, the percent yield of silver is 86.96 %