Animated Solution for Chemistry - Basic Concepts in Chemistry: What volume of hydrogen gas, at $273\text{ K}$ and $1\text{ atm}$ pressure will be consumed in obtaining $21.6\text{ g}$ of elemental boron (atomic mass $=10.8$) from the reduction of boron trichloride by hydrogen?

The Stoichiometry of Boron Reduction: A Journey from Mass to Volume

Imagine you are a chemical engineer tasked with producing pure elemental boron for advanced materials. You have boron trichloride and hydrogen gas. But how much hydrogen do you actually need? This isn't just a theoretical exercise; it's the heart of chemical manufacturing! In this problem, we will bridge the gap between the solid mass of a product and the gaseous volume of a reactant using the elegant principles of stoichiometry and the Ideal Gas Law.

The first step in any chemical journey is to understand the transformation. We are reducing boron trichloride (${\text{BCl}}_3$) using hydrogen gas (${\text{H}}_2$) to obtain elemental boron ($\text{B}$) and hydrogen chloride ($\text{HCl}$).

Before we crunch any numbers, we must speak the language of chemistry: the balanced equation.

This equation is our master recipe. It tells us that to produce exactly $2\text{ moles}$ of boron, we must consume exactly $3\text{ moles}$ of hydrogen gas. This $2:3$ ratio is the golden key to unlocking the problem.

We aren't given the moles of boron directly; instead, we are given its mass: $21.6\text{ g}$. In chemistry, mass is just an illusion; moles are the true currency.

Let's convert this mass into moles using the atomic mass of boron ($10.8\text{ g/mol}$).

What a beautiful coincidence! We need exactly $2\text{ moles}$ of boron, which perfectly matches the stoichiometric coefficient in our balanced equation.

Since the production of $2\text{ moles}$ of boron requires exactly $3\text{ moles}$ of hydrogen gas, we now know that we need $n_{{\text{H}}_2}=3\text{ mol}$.

But the question doesn't ask for moles; it asks for the volume of hydrogen gas at a specific temperature ($273\text{ K}$) and pressure ($1\text{ atm}$). Enter the Ideal Gas Law:

Let's rearrange the formula to solve for volume ($V$) and substitute our known values. We will use the universal gas constant $R=0.0821{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$.

Did you notice something special about the conditions? $273\text{ K}$ and $1\text{ atm}$ are the exact conditions for Standard Temperature and Pressure (STP).

At STP, any ideal gas occupies a molar volume of $22.4\text{ L/mol}$. We could have bypassed the Ideal Gas Law entirely!

Both paths lead us to the same destination. The volume of hydrogen gas required is $67.2\text{ L}$.