Question

Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing 25.00 g of AgNO3 are combined, 13.32 grams of silver are produced. What is the percent yield?

Answer 1

The percent yield can be calculated by dividing the actual yield by the theoretical yield and multiplying by 100. In this case, the percent yield is 150.7%.

Step-by-step explanation:

The percent yield can be calculated by taking the actual yield and dividing it by the theoretical yield, then multiplying by 100. In this reaction, 5.00 g of zinc and 25.00 g of silver nitrate were used to produce 13.32 g of silver. The theoretical yield can be calculated by converting the mass of zinc to moles, using the stoichiometry of the reaction to determine the moles of silver produced, and then converting that back to mass.

Using the molar masses of zinc (65.39 g/mol) and silver nitrate (169.87 g/mol), we can calculate that the theoretical yield of silver is 8.83 g. The percent yield is then calculated as (13.32 g / 8.83 g) * 100 = 150.7%.

Answer 2

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of
`AgNO_3` = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of
`AgNO_3` = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and
`AgNO_3`.

`\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=(5.00g)/(65.38g/mole)=0.0765moles`

`\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=(25.00g)/(168.97g/mole)=0.1479moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 2 mole of
`AgNO_3` react with 1 mole of
`Zn`

So, 0.1479 moles of
`AgNO_3` react with
`(0.1479)/(2)=0.07395` moles of
`Zn`

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

Now we have to calculate the moles of
`Ag`

From the reaction, we conclude that

As, 2 mole of
`AgNO_3` react to give 2 mole of
`Ag`

So, 0.1479 moles of
`AgNO_3` react to give 0.1479 moles of
`Ag`

Now we have to calculate the mass of
`Ag`

`\text{ Mass of }Ag=\text{ Moles of }Ag* \text{ Molar mass of }Ag`

`\text{ Mass of }Ag=(0.1479moles)* (107.87g/mole)=15.95g`

Theoretical yield of
`Ag` = 15.95 g

Experimental yield of
`Ag` = 13.32 g

Now we have to calculate the percent yield.

`\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}* 100`

`\% \text{ yield}=(13.32g)/(15.95g)* 100=83.51\%`

Therefore, the percent yield is, 83.51 %