What is the maximum amount in moles of $P_2O_5$ that can theoretically be made from 186 g of $P_4$ and excess oxygen?

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Answer 1 · 2018-06-28T16:49:42

Well, we need some stoichiometry...

$1/2P_4(s) + 5/2O_2(g) rarr P_2O_5(s)$

$"Moles of phosphorus"$ $=$ $(186*g)/(31.00*g*mol^-1)=6*mol*"phosphorus atoms"...$ ...

And so we can make AT MOST $3*mol$ $P_2O_5$ ...

We need $15*mol*"oxygen ATOMS"$ for equivalence...

$15*molxx16.00*g*mol^-1=240*g$ ...

And AT MOST we can make $"THREE MOLES"$ of $P_2O_5$ ...a mass of $3*molxx112*g*mol^-1=336*mol$ ..

Of course, $P_2O_5$ , is actually $P_4O_10$ ..but we can use this formulation for determination of equivalence...

What is the maximum amount in moles of P_2O_5P_2O_5 that can theoretically be made from 186 g of P_4P_4 and excess oxygen?