Question

Which of the following should occur when placing a certain amount of tetraphosphorus solid in the same container as dihydrogen gas, if `"8.0 mols"` of `"H"_2(g)` are present?

`A)` `"1 mol"` of `"P"_4(s)` reacts.
`B)` `"32 mols"` of `"P"_4(s)` are produced.
`C)` No reaction occurs with `"H"_2(g)`.
`D)` `"16 mols"` of `"PH"_3(g)` is produced.

Answer 1

Solution

Explanation:

B is invalid as 8mol of `H_2` cannot produce 16 mol of `PH_3`

C is not possible as we assume the reactants are at a condition where the reaction is possible

D is also not valid as by B

A is possible if we take `P_4` to be limiting reagent and the `H_2` to be in excess. So 8 mol of `H_2` will react with 1mol of `P_4` and give 4 mol of `PH_3` and 2 mol of unreacted `H_2`.

Answer 2

From the balanced reaction:

`"P"_4(s) + 6"H"_2(g) -> 4"PH"_3(g)`,

we begin with `"8.0 mols"` of `"H"_2(g)`.

Since we have `"6 mols H"_2` for every `"1 mol P"_4(s)`, we know that

`8.0 cancel("mols H"_2(g)) xx ("1 mol P"_4(s))/(6 cancel("mols H"_2(g)))`

`= ul("1.33 mols P"_4(s))`

will react with it. Likewise, since `"P"_4(s)` is `1:4` with `"PH"_3`, we would produce `ul("5.33 mols")` of `"PH"_3(g)`.

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`A)` is wrong, as we have shown it reacts with more than `"1 mol"` of `"P"_4(s)`... It is the only answer that could potentially have been right, if we started with `"6.0 mols H"_2`... but we didn't!

`B)` is wrong, because that is far too many mols produced.

`C)` Phosphorus is clearly reactive towards `"H"_2(g)`. It looks like this:

![)

The bond angles (`60^@`!) are strained, making `"P"_4(s)` quite reactive...

`D)` is wrong, just as `B)` is also wrong. We found only `"5.33 mols"` produced...