Respuesta :
Answer : The mass of [tex]PH_3[/tex] produced from the reaction is, 0.651 grams.
Explanation :
The chemical reactions used are:
(1) [tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
(2) [tex]6NO+P_4\rightarrow P_4O_6+3N_2[/tex]
(3) [tex]P_4O_6+6H_2O\rightarrow 4H_3PO_4[/tex]
(4) [tex]4H_3PO_4\rightarrow PH_3+3H_3PO_4[/tex]
First we have to calculate the moles of [tex]NH_3[/tex]
[tex]\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}[/tex]
Molar mass of [tex]NH_3[/tex] = 17 g/mol
[tex]\text{Moles of }NH_3=\frac{1.95g}{17g/mol}=0.115mol[/tex]
Now we have to calculate the moles of [tex]NO[/tex]
From the balanced chemical reaction 1, we conclude that:
As, 4 moles of [tex]NH_3[/tex] react to give 4 moles of [tex]NO[/tex]
So, 0.115 moles of [tex]NH_3[/tex] react to give 0.115 moles of [tex]NO[/tex]
Now we have to calculate the moles of [tex]P_4O_6[/tex]
From the balanced chemical reaction 2, we conclude that:
As, 6 moles of [tex]NO[/tex] react to give 1 moles of [tex]P_4O_6[/tex]
So, 0.115 moles of [tex]NO[/tex] react to give [tex]\frac{0.115}{6}=0.0192[/tex] moles of [tex]P_4O_6[/tex]
Now we have to calculate the moles of [tex]H_3PO_4[/tex]
From the balanced chemical reaction 3, we conclude that:
As, 1 moles of [tex]P_4O_6[/tex] react to give 4 moles of [tex]H_3PO_4[/tex]
So, 0.0192 moles of [tex]P_4O_6[/tex] react to give [tex]0.0192\times 4=0.0768[/tex] moles of [tex]H_3PO_4[/tex]
Now we have to calculate the moles of [tex]PH_3[/tex]
From the balanced chemical reaction 4, we conclude that:
As, 4 moles of [tex]H_3PO_4[/tex] react to give 1 moles of [tex]PH_3[/tex]
So, 0.0768 moles of [tex]H_3PO_4[/tex] react to give [tex]\frac{0.0768}{4}=0.0192[/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]
Molar mass of [tex]PH_3[/tex] = 33.9 g/mole
[tex]\text{ Mass of }PH_3=(0.0192moles)\times (33.9g/mole)=0.651g[/tex]
Therefore, the mass of [tex]PH_3[/tex] produced from the reaction is, 0.651 grams.
