Question

How many grams of phosphine (PH3) can form when 11.3 g of phosphorus and 98.2 L of hydrogen gas react at STP?

Answer 1

Approximately `12.4` grams of phosphine will be produced.

Explanation:

Start by writing a balanced chemical equation. Here is the unbalanced:

`P_4 + H_2 -> PH_3`

Comparing coefficients to balance, we get:

`P_4 + 6H_2 -> 4PH_3`

Now notice that they give the amount of hydrogen gas in litres. That will have to be converted to moles. We can use the molar volume of gas, `22.4 L/"mol"` to do this.

`98.2 L * "mol"/(22.4 L) = 4.38" "mol`

Now we have to convert grams of phosphorus to moles of phosphorus. Since phosphorus is `P_4`, its molar mass will be `4(30.97) = 123.88 g/"mol"`.

`11.3 g * "mol"/(123.88 g) = 0.0912" "mol" "P_4`

Clearly this will run out much before the hydrogen gas, so we can only react this much, therefore, the amount of phosphine produced will be `4(0.0912) = 0.3648` moles. The final step is to convert this into grams. We can check via the periodic table that the molar mass of phosphine is `34.0` g/`"mol"`.

`0.3648" "mol * 34.0 g/"mol" = 12.4" "g" phosphine"`

Hopefully this helps!