Question

\(2 \mathrm{~g}\) of aluminium is treated separately with excess of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and excess of \(\mathrm{NaOH}\). The ratio of the volumes of hydrogen evolved is (a) \(1: 1\) (b) \(1: 2\) (c) \(2: 1\) (d) \(2: 3\)

Short answer

The ratio of volumes of hydrogen evolved is (a) 1:1.

Step-by-step solution

Identify the reactions involved

First, understand the chemical reactions that occur when aluminum reacts with dilute sulfuric acid and sodium hydroxide. The reactions are: \[ 2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2 \] and \[ 2Al + 2NaOH + 6H_2O \rightarrow 2NaAl(OH)_4 + 3H_2 \].

Molar mass calculation

The molar mass of aluminum, \(Al\), is approximately \(27 \, \text{g/mol}\). Calculate the number of moles in \(2 \, \text{g}\) of aluminum: \[ \text{Moles of Al} = \frac{2}{27} \approx 0.0741 \, \text{mol} \].

Volume of hydrogen calculation for H2SO4 reaction

From the balanced equation for \(\text{H}_2\text{SO}_4\), \(2 \, \text{moles}\) of \(Al\) produce \(3 \, \text{moles}\) of \(H_2\). Therefore, \(0.0741 \, \text{moles}\) of \(Al\) produces: \[ \frac{3}{2} \times 0.0741 = 0.1112 \, \text{mol of } H_2 \].

Volume of hydrogen calculation for NaOH reaction

From the balanced equation with \(\text{NaOH}\), the same \(2 \, \text{moles}\) of \(Al\) produce \(3 \, \text{moles}\) of \(H_2\), giving the same number of moles of \(0.1112 \, \text{mol}\).

Consistency check

Both reactions yield \(0.1112 \, \text{mol of } H_2\). Hence, the ratio of hydrogen evolved is equal. Since 1 mole of any gas occupies the same volume at a given temperature and pressure, the volumes are equal.

Key concepts

Balanced Chemical Equations

Balanced chemical equations are crucial when studying chemical reactions, especially in stoichiometry. When aluminum reacts with sulfuric acid or a base like sodium hydroxide, balancing the equation ensures we know exactly how much of each substance is involved. For aluminum reacting with sulfuric acid, the balanced equation is:\[ 2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2 \]For the reaction of aluminum with sodium hydroxide, the balanced equation is:\[ 2Al + 2NaOH + 6H_2O \rightarrow 2NaAl(OH)_4 + 3H_2 \]In both cases, the coefficients of each compound tell us the stoichiometric ratios necessary for the reaction. This helps us determine how much hydrogen gas will be evolved. A balanced equation reflects the conservation of mass, ensuring that the number of atoms for each element is the same on both sides of the equation.

Molar Mass Calculation

Calculating molar mass is essential in converting mass to moles. The molar mass of Al is around \(27 \, \text{g/mol}\). Using this, we can determine how many moles are present in a given mass. For example, if we have \(2 \, \text{g}\) of aluminum, we calculate:

• Number of moles = \( \frac{2}{27} \approx 0.0741 \, \text{mol} \)

Knowing the moles is useful for comparing with the balanced equations to find the amounts of other substances produced or consumed in the reactions. Molar mass calculations link the practical mass measurements we make to the moles which appear in stoichiometric calculations.

Volume of Gas Calculation

Understanding how to calculate the volume of a gas is crucial, as gases are common products in chemical reactions. In the context of this problem, hydrogen gas is produced. At standard temperature and pressure (STP), one mole of any gas occupies approximately \(22.4 \, \text{liters}\). From the balanced equations:

• Al + \(H_2SO_4:\) \(0.0741 \, \text{mol of Al} \rightarrow 0.1112 \, \text{mol of } H_2 \)
• Al + \(NaOH:\) \(0.0741 \, \text{mol of Al} \rightarrow 0.1112 \, \text{mol of } H_2 \)

To find the actual volumes, we multiply the amount of moles of hydrogen by \(22.4 \, \text{L/mol}\). In both reactions, the expressed molar ratio leads to the same volume of hydrogen gas. Therefore, the volume ratio in the reactions is \(1:1\) as both produce equal moles of \(H_2\).

Reaction of Aluminum with Acids and Bases

Aluminum is a metal that reacts with acids and bases, often producing hydrogen gas in the process. When aluminum reacts with \(H_2SO_4\) (an acid), it forms aluminum sulfate and hydrogen gas. Similarly, in a reaction with \(NaOH\) (a base), aluminum produces sodium aluminate and hydrogen gas. These reactions show that aluminum is amphoteric, meaning it can react with both acids and bases.

• Reaction with \(H_2SO_4\): Produces \(Al_2(SO_4)_3\) and \(H_2\), exhibiting typical metal-acid behavior.
• Reaction with \(NaOH\): Produces \(NaAl(OH)_4\) along with \(H_2\), showing its ability to dissolve in bases.

The evolution of hydrogen gas is a common trait in both reactions. Understanding these reactions is key in explaining the amphoteric nature of aluminum and its industrial applications, such as in water treatment and as a component in chemical manufacturing.