Answer :
Answer: The amount of [tex]P_4O_{10}[/tex] formed is 469.8 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of phosphine = 225 g
Molar mass of phosphine = 34 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol[/tex]
The given chemical reaction follows:
[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]
Assuming that oxygen gas is present in excess, it is considered as an excess reagent.
Phosphine is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
4 moles of phosphine produces 1 mole of [tex]P_4O_{10}[/tex]
So, 6.62 moles of phosphine will produce = [tex]\frac{1}{4}\times 6.62=1.655mol[/tex] of [tex]P_4O_{10}[/tex]
Now, calculating the mass of [tex]P_4O_{10}[/tex] by using equation 1:
Molar mass of [tex]P_4O_{10}[/tex] = 283.9 g/mol
Moles of [tex]P_4O_{10}[/tex] = 1.655 moles
Putting values in equation 1, we get:
[tex]1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g[/tex]
Hence, the amount of [tex]P_4O_{10}[/tex] formed is 469.8 grams.