Question

How many grams of phosphine ($P{H}_3$) can form when 11.3 g of phosphorus and 98.2 L of hydrogen gas reacts at STP?

Answer 1

Hint: First, you have to convert the volume of the hydrogen gas to a number of moles and then the given mass of phosphorus to the number of moles. The chemical equation involved in the question will be:
$P_4+H_2\to P{H}_3$
Balance the equation to solve the question.

Complete step-by-step answer:Let us first write the equation of the formation of phosphine from phosphorus and hydrogen. The elemental formula of phosphorus is $P_4$ and hydrogen is $H_2$. The reaction is:
$P_4+H_2\to P{H}_3$
Now, let us balance this reaction. The balanced equation will be:
$P_4+6H_2\to 4P{H}_3$
The given value of hydrogen is in liters, i.e., 98.2 L. So, we have to convert it into a number of moles. We know that the volume of the gas at STP is 22.4 L /mol. We can divide the given volume of the hydrogen gas by the volume at STP to get the number of moles:
${\displaystyle \frac{98.2}{22.4}}=4.38\text{ moles}$
So, the mole of hydrogen is 4.38.
The given value of phosphorus is in grams, so it also has to be converted into a number of moles. The molecular mass of phosphorus ($P_4$) is 123.88 g /mol. The number of moles will be:
${\displaystyle \frac{11.3}{123.88}}=0.0912\text{ moles}$
We can clearly see that the moles of phosphorus are very less, so it will act as a limiting reagent. So, the phosphine will be produced until the phosphorus is there. The phosphine formed will be:
$4\text{ x 0}\text{.0912=0}\text{.3648 mol}$
We have the number of moles phosphine formed, hence the amount can be calculated as:
$0.3648\text{ x 34 = 12}\text{.4 g}$
12.4 grams of phosphine will be formed.

Note:The molecular mass of phosphine ($P{H}_3$) is calculated as: 31 + 1 + 1 + 1 = 34 gram /mol. The number of moles is calculated as the ratio of the given mass of the substance by the molecular mass of the substance.