Respuesta :
Answer : The percent yield of [tex]P_4O_{10}[/tex] is, 87.7 %
Solution : Given,
Moles of [tex]P_4[/tex] = 0.200 mole
Moles of [tex]O_2[/tex] = 0.200 mole
Molar mass of [tex]P_4O_{10}[/tex] = 283.9 g/mole
First we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]P_4+5O_2\rightarrow P_4O_{10}[/tex]
From the balanced reaction we conclude that
As, 5 mole of [tex]O_2[/tex] react with 1 mole of [tex]P_4[/tex]
So, 0.200 moles of [tex]O_2[/tex] react with [tex]\frac{0.200}{5}=0.04[/tex] moles of [tex]P_4[/tex]
From this we conclude that, [tex]P_4[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]P_4O_{10}[/tex]
From the reaction, we conclude that
As, 5 mole of [tex]O_2[/tex] react to give 1 mole of [tex]P_4O_{10}[/tex]
So, 0.200 moles of [tex]O_2[/tex] react to give [tex]\frac{0.200}{5}=0.04[/tex] moles of [tex]P_4O_{10}[/tex]
Now we have to calculate the mass of [tex]P_4O_{10}[/tex]
[tex]\text{ Mass of }P_4O_{10}=\text{ Moles of }P_4O_{10}\times \text{ Molar mass of }P_4O_{10}[/tex]
[tex]\text{ Mass of }P_4O_{10}=(0.04moles)\times (283.9g/mole)=11.4g[/tex]
Theoretical yield of [tex]P_4O_{10}[/tex] = 11.4 g
Experimental yield of [tex]P_4O_{10}[/tex] = 10.0 g
Now we have to calculate the percent yield of [tex]P_4O_{10}[/tex]
[tex]\% \text{ yield of }P_4O_{10}=\frac{\text{ Experimental yield of }P_4O_{10}}{\text{ Theretical yield of }P_4O_{10}}\times 100[/tex]
[tex]\% \text{ yield of }P_4O_{10}=\frac{10.0g}{11.4g}\times 100=87.7\%[/tex]
Therefore, the percent yield of [tex]P_4O_{10}[/tex] is, 87.7 %
