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Shkiper50 [21]
3 years ago
5

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?
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
6 0

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}=\frac{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}=\frac{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 \frac{1}{2}\times 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 \frac{3}{2}\times 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\times \text{ Molar mass of }K_2CO_3

\text{ Mass of }K_2CO_3=(1.0599moles)\times (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}}\times 100

\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%

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

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