Respuesta :
Answer : The percent yield of [tex]P_2O_5[/tex] is, 30.39 %
Solution : Given,
Mass of P = 2.80 g
Molar mass of P = 31 g/mole
Molar mass of [tex]P_2O_5[/tex] = 284 g/mole
First we have to calculate the moles of P.
[tex]\text{ Moles of }P=\frac{\text{ Mass of }P}{\text{ Molar mass of }P}=\frac{2.80g}{31g/mole}=0.0903moles[/tex]
Now we have to calculate the moles of [tex]NH_3[/tex]
The balanced chemical reaction is,
[tex]4P+5O_2\rightarrow 2P_2O_5[/tex]
From the reaction, we conclude that
As, 4 mole of [tex]P[/tex] react to give 2 mole of [tex]P_2O_5[/tex]
So, 0.0903 moles of [tex]P[/tex] react to give [tex]\frac{0.0903}{4}\times 2=0.04515[/tex] moles of [tex]P_2O_5[/tex]
Now we have to calculate the mass of [tex]P_2O_5[/tex]
[tex]\text{ Mass of }P_2O_5=\text{ Moles of }P_2O_5\times \text{ Molar mass of }P_2O_5[/tex]
[tex]\text{ Mass of }P_2O_5=(0.04515moles)\times (284g/mole)=12.8g[/tex]
Theoretical yield of [tex]P_2O_5[/tex] = 12.8 g
Experimental yield of [tex]P_2O_5[/tex] = 3.89 g
Now we have to calculate the percent yield of [tex]P_2O_5[/tex]
[tex]\% \text{ yield of }P_2O_5=\frac{\text{ Experimental yield of }P_2O_5}{\text{ Theoretical yield of }P_2O_5}\times 100[/tex]
[tex]\% \text{ yield of }P_2O_5=\frac{3.89g}{12.8g}\times 100=30.39\%[/tex]
Therefore, the percent yield of [tex]P_2O_5[/tex] is, 30.39 %
