# 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) $\text{1 mol}$ of ${\text{P}}_{4} \left(s\right)$ reacts. B) $\text{32 mols}$ of ${\text{P}}_{4} \left(s\right)$ are produced. C) No reaction occurs with ${\text{H}}_{2} \left(g\right)$. D) $\text{16 mols}$ of ${\text{PH}}_{3} \left(g\right)$ is produced.

Aug 21, 2017

Solution

#### Explanation:

B is invalid as 8mol of ${H}_{2}$ cannot produce 16 mol of $P {H}_{3}$

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 $P {H}_{3}$ and 2 mol of unreacted ${H}_{2}$.

Aug 21, 2017

From the balanced reaction:

${\text{P"_4(s) + 6"H"_2(g) -> 4"PH}}_{3} \left(g\right)$,

we begin with $\text{8.0 mols}$ of ${\text{H}}_{2} \left(g\right)$.

Since we have ${\text{6 mols H}}_{2}$ for every ${\text{1 mol P}}_{4} \left(s\right)$, we know that

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

$= \underline{{\text{1.33 mols P}}_{4} \left(s\right)}$

will react with it. Likewise, since ${\text{P}}_{4} \left(s\right)$ is $1 : 4$ with ${\text{PH}}_{3}$, we would produce $\underline{\text{5.33 mols}}$ of ${\text{PH}}_{3} \left(g\right)$.

A) is wrong, as we have shown it reacts with more than $\text{1 mol}$ of ${\text{P}}_{4} \left(s\right)$... It is the only answer that could potentially have been right, if we started with ${\text{6.0 mols H}}_{2}$... but we didn't!

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

C) Phosphorus is clearly reactive towards ${\text{H}}_{2} \left(g\right)$. It looks like this:

The bond angles (${60}^{\circ}$!) are strained, making ${\text{P}}_{4} \left(s\right)$ quite reactive...

D) is wrong, just as B) is also wrong. We found only $\text{5.33 mols}$ produced...