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

Chapter 11, Problem 114P

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QUESTION:

Consider the reaction:

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

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

(b) If 48.3 L of $${P H}_{3}$$, measured at STP, forms, what is the percent yield?

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 $$\mathrm{P}_{4}$$ :

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

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