Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hy

Q: Why do we need to find the limiting reagent?

In any chemical reaction with multiple reactants, one reactant will usually run out first. This reactant 'limits' the amount of product that can be formed. If we use the excess reactant for our calculations, we will overestimate the product!

Q: How did we get the moles of sulfuric acid?

For solutions, moles can be calculated by multiplying the molarity (M) by the volume in liters (V). Here, $5.0 \text{ M} \times 0.050 \text{ L} = 0.25 \text{ moles}$.

Q: Why is the stoichiometric ratio of H2SO4 to H2 1:1?

Look at the balanced equation: $2\text{Al} + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3\text{H}_2$. The coefficients for both $\text{H}_2\text{SO}_4$ and $\text{H}_2$ are 3. Therefore, the ratio is 3:3, which simplifies to 1:1.

Q: What if aluminium was the limiting reagent?

If aluminium were the limiting reagent, we would use its moles to find the moles of hydrogen. Since the ratio of Al to $\text{H}_2$ is 2:3, we would multiply the moles of Al by $3/2$ to get the moles of $\text{H}_2$.

Q: Can I use $R = 8.314$ instead of $0.082$?

The value of the universal gas constant $R$ depends on the units of pressure and volume. Since the pressure is given in atm and we need the volume in liters, we must use $R = 0.082 \text{ atm L mol}^{-1} \text{ K}^{-1}$. Using $8.314$ would require pressure in Pascals and yield volume in cubic meters.

Question

Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters (L) produced at $300	ext{ K}$ and $1.0	ext{ atm}$ pressure, when $5.4	ext{ g}$ of aluminium and $50.0	ext{ mL}$ of $5.0	ext{ M}$ sulfuric acid are combined for the reaction?
(Use molar mass of aluminium as $27.0	ext{ g mol}^{-1}$, $R = 0.082	ext{ atm L mol}^{-1}	ext{ K}^{-1}$)

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Reaction

Imagine a laboratory beaker sitting on a bench. Inside, we have a clear solution of sulfuric acid ($5.0	ext{ M}$). We drop in a few shiny pieces of solid aluminium ($5.4	ext{ g}$).

Instantly, the mixture starts to fizz! Bubbles of hydrogen gas rapidly rise to the surface. But how much gas is actually being produced? To answer this, we need to translate this physical reality into the language of chemistry.

The foundation of any stoichiometry problem is a perfectly balanced chemical equation. For this reaction, aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen gas.

The balanced equation is:

$2	ext{Al} + 3	ext{H}_2	ext{SO}_4 ightarrow 	ext{Al}_2(	ext{SO}_4)_3 + 3	ext{H}_2$

This tells us the fundamental recipe: **two moles of aluminium require exactly three moles of sulfuric acid** to react completely.

### The Mole Hunt: Quantifying the Reactants

Before we can determine the amount of product, we must find out exactly how much of each reactant we have in our beaker. We need to convert the given masses and volumes into moles.

For the solid aluminium, we use its mass and molar mass.

$n_{	ext{Al}} = \frac{5.4	ext{ g}}{27.0	ext{ g mol}^{-1}} = 0.2	ext{ moles}$

For the sulfuric acid solution, we use its molarity and volume. Remember to convert the volume from milliliters to liters!

$n_{	ext{H}_2	ext{SO}_4} = 5.0	ext{ M} 	imes 0.050	ext{ L} = 0.25	ext{ moles}$

Now we have our starting inventory: $0.2	ext{ moles}$ of $	ext{Al}$ and $0.25	ext{ moles}$ of $	ext{H}_2	ext{SO}_4$.

### The Bottleneck: Finding the Limiting Reagent

Here is where many students make a critical error. You cannot simply pick one reactant and use it to find the product. We must identify the **limiting reagent**—the reactant that will run out first and stop the reaction.

To find the limiting reagent, we divide the available moles of each reactant by its stoichiometric coefficient from the balanced equation.

For Aluminium:

$\frac{0.2}{2} = 0.1$

For Sulfuric Acid:

$\frac{0.25}{3} \approx 0.083$

Since $0.083$ is less than $0.1$, **sulfuric acid is our limiting reagent**. It will be completely consumed, while some aluminium will be left over unreacted in the beaker.

### The Gas Law: From Moles to Volume

Now that we know sulfuric acid dictates the reaction, we use it to find the moles of hydrogen gas produced.

Looking back at our balanced equation, the stoichiometric ratio between $	ext{H}_2	ext{SO}_4$ and $	ext{H}_2$ is $3:3$, which simplifies to a beautiful $1:1$ ratio.

Therefore, the moles of hydrogen gas produced will be exactly equal to the moles of sulfuric acid consumed.

$n_{	ext{H}_2} = 0.25	ext{ moles}$

Finally, the question asks for the volume of this gas at a specific temperature ($300	ext{ K}$) and pressure ($1.0	ext{ atm}$). This is a classic application of the **Ideal Gas Law**.

$PV = nRT$

Rearranging for volume:

$V = \frac{nRT}{P}$

Substituting our values:

$V = \frac{0.25 	imes 0.082 	imes 300}{1.0}$

$V = 0.25 	imes 24.6 = 6.15	ext{ L}$

The reaction will produce exactly $6.15	ext{ liters}$ of hydrogen gas. A perfect harmony of stoichiometry and gas laws!

Enter Numerical Value:

Visualized Solution

$2 Al + 3 H_{2} SO_{4} \to Al_{2} (SO_{4})_{3} + 3 H_{2}$

$Initial Moles of Reactants$

$Identifying the Limiting Reagent$

$Moles of Hydrogen Gas$

$Applying Ideal Gas Law$

$Final Volume Calculation$

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Setup

The Mole Hunt

The Bottleneck

The Gas Law

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2Al+3H2​SO4​→Al2​(SO4​)3​+3H2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Moles of Reactants

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Identifying the Limiting Reagent

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Hydrogen Gas

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Ideal Gas Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Volume Calculation