Question

Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate and 90 g of Aluminum Carbonate, what is my theoretical yeild in grams of Potassium Carbonate? If the reaction results in 125 g of potassium carbonate, what is my percent yeild?

Answer 1

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of
`K_3PO_4` = 150 g

Mass of
`Al_2(CO_3)_3` = 90 g

Molar mass of
`K_3PO_4` = 212.27 g/mole

Molar mass of
`Al_2(CO_3)_3` = 233.99 g/mole

Molar mass of
`K_2CO_3` = 138.205 g/mole

First we have to calculate the moles of
`K_3PO_4` and
`Al_2(CO_3)_3`

`\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=(150g)/(212.27g/mole)=0.7066moles`

`\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=(90g)/(233.99g/mole)=0.3846moles`

The given balanced reaction is,

`2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4`

From the given reaction, we conclude that

2 moles of
`K_3PO_4` react with 1 mole of
`Al_2(CO_3)_3`

0.7066 moles of
`K_3PO_4` react with
`(1)/(2)* 0.7066=0.3533` moles of
`Al_2(CO_3)_3`

But the moles of
`Al_2(CO_3)_3` is, 0.3846 moles.

So,
`Al_2(CO_3)_3` is an excess reagent and
`K_3PO_4` is a limiting reagent.

Now we have to calculate the moles of
`K_2CO_3`.

As, 2 moles of
`K_3PO_4` react to give 3 moles of
`K_2CO_3`

So, 0.7066 moles of
`K_3PO_4` react to give
`(3)/(2)* 0.7066=1.0599` moles of
`K_2CO_3`

Now we have to calculate the mass of
`K_2CO_3`.

`\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3* \text{ Molar mass of }K_2CO_3`

`\text{ Mass of }K_2CO_3=(1.0599moles)* (138.205g/mole)=146.483g`

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

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

`\% \text{ yield of }K_2CO_3=(125g)/(146.483g)* 100=85.33\%`

Therefore, the % yield of potassium carbonate is, 85.33%