Consider the reaction:P4(s) + 6H2(g) ? 4 PH3(g)(a) If 88.6 L of H2(g), measured at STP

Step 1 of 4

Given reaction:

\(\mathrm{P}_{4}(\mathrm{s})+6 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PH}_{3}(\mathrm{g})\)

(a) If 88.6 L of \(\mathrm{H}_{2}\) (g), measured at STP, is allowed to react with 158.3 g of \(\mathrm{P}_{4}\) What is the limiting reactant?

To find the limiting reactant, we should know the moles of each reactant. So let’s find the number of moles of each reactant :

First, let’s find number of moles of \(\mathrm{H}_{2}\) :

We know, at STP(Standard Temperature and Pressure), one mole of any gas occupies 22.4 L.

\(\begin{array}{l}\text{Moles of}\\ \begin{aligned}\mathrm{H}_{2} & =\frac{88.6 \mathrm{L}}{22.4 \mathrm{L} / \mathrm{mol}} \\ & =3.955 \text { moles }\end{aligned}\end{array}\)

Now, let’s find number of moles of \(\mathrm{P}_{4}\) :

\(\begin{array}{l}\text{Molar mass of } P_{4}=123.89 \mathrm{g} / \mathrm{mol} \text{.}\\ \text { moles of } \mathrm{P}_{4}=\frac{158.3 \mathrm{g}}{123.89 \mathrm{g} / \mathrm{mol}} \\ =1.277 \text { moles } \text {. } \\ \end{array}\)