Respuesta :
Answer : The maximum number of grams of [tex]PH_3[/tex] formed is, 8.955 g
Solution : Given,
Mass of phosphorous = 8.2 g
Mass of hydrogen = 4 g
Molar mass of [tex]P_4[/tex] = 123.6 g/mole
Molar mass of [tex]H_2[/tex] = 2.016 g/mole
Molar mass of [tex]PH_3[/tex] = 33.924 g/mole
The balanced chemical reaction is,
[tex]P_4(s)+6H_2(g)\rightarrow 4PH_3(g)[/tex]
First we have to calculate the moles [tex]P_4[/tex] and [tex]H_2[/tex]
[tex]\text{ Moles of }P_4=\frac{\text{ Mass of }P_4}{\text{ Molar mass of }P_4}=\frac{8.2g}{123.6g/mole}=0.066moles[/tex]
[tex]\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{4g}{2.016g/mole}=1.98moles[/tex]
From the reaction, we conclude that
1 mole of [tex]P_4[/tex] react with 6 moles of [tex]H_2[/tex]
0.066 moles of [tex]P_4[/tex] react with [tex]6\times 0.066=0.396[/tex] moles of [tex]H_2[/tex]
That means the [tex]H_2[/tex] is in excess amount and [tex]P_4[/tex] is in limited amount.
Now we have to calculate the moles of [tex]PH_3[/tex].
As, 1 mole of [tex]P_4[/tex] react to give 4 moles of [tex]PH_3[/tex]
So, 0.066 moles of [tex]P_4[/tex] react to give [tex]4\times 0.066=0.264[/tex] moles of [tex]PH_3[/tex]
Now we have to calculate the mass of [tex]PH_3[/tex]
[tex]\text{ Mass of }PH_3=\text{ Moles of }PH_3\times \text{ Molar mass of }PH_3[/tex]
[tex]\text{ Mass of }PH_3=(0.264moles)\times (33.924g/mole)=8.955g[/tex]
Therefore, the maximum number of grams of [tex]PH_3[/tex] formed is, 8.955 g
