Respuesta :
Answer: The mass of [tex]H_3PO_4[/tex] produced is, 13.82 grams.
Explanation : Given,
Mass of [tex]P_4O_{10}[/tex] = 10.00 g
Mass of [tex]H_2O[/tex] = 12.00 g
Molar mass of [tex]P_4O_{10}[/tex] = 283.89 g/mol
Molar mass of [tex]H_2O[/tex] = 18 g/mol
First we have to calculate the moles of [tex]P_4O_{10}[/tex] and [tex]H_2O[/tex].
[tex]\text{Moles of }P_4O_{10}=\frac{\text{Given mass }P_4O_{10}}{\text{Molar mass }P_4O_{10}}[/tex]
[tex]\text{Moles of }P_4O_{10}=\frac{10.0g}{283.89g/mol}=0.0352mol[/tex]
and,
[tex]\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}[/tex]
[tex]\text{Moles of }H_2O=\frac{12.0g}{18g/mol}=0.666mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]P_4O_{10}+6H_2O\rightarrow 4H_3PO_4[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]P_4O_{10}[/tex] react with 6 mole of [tex]H_2O[/tex]
So, 0.0352 moles of [tex]P_4O_{10}[/tex] react with [tex]0.0352\times 6=0.211[/tex] moles of [tex]H_2O[/tex]
From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]P_4O_{10}[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]H_3PO_4[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]P_4O_{10}[/tex] react to give 4 mole of [tex]H_3PO_4[/tex]
So, 0.0352 mole of [tex]P_4O_{10}[/tex] react to give [tex]0.0352\times 4=0.141[/tex] mole of [tex]H_3PO_4[/tex]
Now we have to calculate the mass of [tex]H_3PO_4[/tex]
[tex]\text{ Mass of }H_3PO_4=\text{ Moles of }H_3PO_4\times \text{ Molar mass of }H_3PO_4[/tex]
Molar mass of [tex]H_3PO_4[/tex] = 98.00 g/mole
[tex]\text{ Mass of }H_3PO_4=(0.141moles)\times (98.00g/mole)=13.82g[/tex]
Therefore, the mass of [tex]H_3PO_4[/tex] produced is, 13.82 grams.
