Respuesta :
Answer : The theoretical yield of potassium carbonate is, 146.483 g
The percent yield of potassium carbonate is, 85.33 %
Solution : Given,
Mass of [tex]K_3PO_4[/tex] = 150 g
Mass of [tex]Al_2(CO_3)_3[/tex] = 90 g
Molar mass of [tex]K_3PO_4[/tex] = 212.27 g/mole
Molar mass of [tex]Al_2(CO_3)_3[/tex] = 233.99 g/mole
Molar mass of [tex]K_2CO_3[/tex] = 138.205 g/mole
First we have to calculate the moles of [tex]K_3PO_4[/tex] and [tex]Al_2(CO_3)_3[/tex]
[tex]\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[/tex]
[tex]\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[/tex]
The given balanced reaction is,
[tex]2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4[/tex]
From the given reaction, we conclude that
2 moles of [tex]K_3PO_4[/tex] react with 1 mole of [tex]Al_2(CO_3)_3[/tex]
0.7066 moles of [tex]K_3PO_4[/tex] react with [tex]\frac{1}{2}\times 0.7066=0.3533[/tex] moles of [tex]Al_2(CO_3)_3[/tex]
But the moles of [tex]Al_2(CO_3)_3[/tex] is, 0.3846 moles.
So, [tex]Al_2(CO_3)_3[/tex] is an excess reagent and [tex]K_3PO_4[/tex] is a limiting reagent.
Now we have to calculate the moles of [tex]K_2CO_3[/tex].
As, 2 moles of [tex]K_3PO_4[/tex] react to give 3 moles of [tex]K_2CO_3[/tex]
So, 0.7066 moles of [tex]K_3PO_4[/tex] react to give [tex]\frac{3}{2}\times 0.7066=1.0599[/tex] moles of [tex]K_2CO_3[/tex]
Now we have to calculate the mass of [tex]K_2CO_3[/tex].
[tex]\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3[/tex]
[tex]\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g[/tex]
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 :
[tex]\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100[/tex]
[tex]\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%[/tex]
Therefore, the % yield of potassium carbonate is, 85.33%
