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P4(s) + 10 Cl2(g) → 4 PCl5(g)
P4 + 5 O2 → 2 P2O5
(186 g P4) / (123.8950 g/mol) x (2/1) x (141.9447 g/mol) = 426 g P2O5
P4(s) + 10 Cl2(g) → 4 PCl5(g)
P4 + 5 O2 → 2 P2O5
(186 g P4) / (123.8950 g/mol) x (2/1) x (141.9447 g/mol) = 426 g P2O5
Answer : The maximum amount in moles of [tex]P_2O_5[/tex] is, 3 moles
Solution : Given,
Mass of [tex]P_4[/tex] = 186 g
Molar mass of [tex]P_4[/tex] = 123.88 g/mole
First we have to calculate the moles of [tex]P_4[/tex].
[tex]\text{Moles of }P_4=\frac{\text{Mass of }P_4}{\text{Molar mass of }P_4}=\frac{186g}{123.88g/mole}=1.50moles[/tex]
Now we have to calculate the moles of [tex]P_2O_5[/tex].
The balanced chemical reaction will be,
[tex]P_4+5O_2\rightarrow 2P_2O_5[/tex]
From the balanced reaction, we conclude that
As, 1 mole of [tex]P_4[/tex] react to give 2 moles of [tex]P_2O_5[/tex]
So, 1.50 moles of [tex]P_4[/tex] react to give [tex]2\times 1.50=3moles[/tex] of [tex]P_2O_5[/tex]
Therefore, the maximum amount in moles of [tex]P_2O_5[/tex] is, 3 moles