3.90 g mixture of al and aluminum oxide when reacted with a solution of sodium hydroxide, produced 840 ml of a gas at NTP. Find the composition of the mixture...

To determine the composition of the mixture of aluminum (Al) and aluminum oxide (Al2O3) that reacted with sodium hydroxide (NaOH) to produce a gas, we can start by analyzing the chemical reactions involved. When aluminum reacts with sodium hydroxide, hydrogen gas (H2) is produced, and aluminum oxide can react under specific conditions, but we will focus on the aluminum for this calculation.

Understanding the Reaction

The reaction between aluminum and sodium hydroxide can be represented as follows:

2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2

In this reaction, two moles of aluminum react with sodium hydroxide to produce hydrogen gas. At Normal Temperature and Pressure (NTP), one mole of gas occupies 22.4 liters. Therefore, we need to calculate how many moles of hydrogen gas were produced from the volume provided.

Calculating Moles of Hydrogen Gas

Given that 840 mL of hydrogen gas was produced, we can convert this volume to liters:

840 mL = 0.840 L

Next, we calculate the number of moles of hydrogen gas using the ideal gas law. At NTP, 1 mole occupies 22.4 L:

Number of moles of H2 = Volume (L) / Molar volume (L/mol) = 0.840 L / 22.4 L/mol = 0.0375 moles

Relating Moles of Aluminum to Hydrogen Production

From the balanced equation, we see that 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, we can set up a ratio to find the amount of aluminum that reacted:

2 moles Al → 3 moles H2

From this, we can calculate the moles of aluminum that produced 0.0375 moles of hydrogen:

(2 moles Al / 3 moles H2) = (x moles Al / 0.0375 moles H2)

Solving for x gives:

x = (2/3) * 0.0375 = 0.0250 moles Al

Calculating Mass of Aluminum

Now, we can convert moles of aluminum to grams. The molar mass of aluminum is approximately 27 g/mol:

Mass of Al = moles × molar mass = 0.0250 moles × 27 g/mol = 0.675 g

Determining Composition of the Mixture

We know the total mass of the mixture is 3.90 g. We have found that the mass of aluminum is 0.675 g. Therefore, we can now find the mass of aluminum oxide in the mixture:

Mass of Al2O3 = Total mass - Mass of Al = 3.90 g - 0.675 g = 3.225 g

Calculating Moles of Aluminum Oxide

Next, we can find the moles of aluminum oxide. The molar mass of Al2O3 is approximately 102 g/mol:

Moles of Al2O3 = Mass / Molar mass = 3.225 g / 102 g/mol ≈ 0.0316 moles

Final Composition Analysis

Now we can summarize the composition of the original mixture:

• Mass of Aluminum: 0.675 g
• Mass of Aluminum Oxide: 3.225 g
• Total Mass: 3.90 g
• Moles of Aluminum: 0.0250 moles
• Moles of Aluminum Oxide: 0.0316 moles

To express this as a percentage composition:

• Percentage of Al = (0.675 g / 3.90 g) × 100 ≈ 17.31%
• Percentage of Al2O3 = (3.225 g / 3.90 g) × 100 ≈ 82.69%

Thus, the mixture consists of approximately 17.31% aluminum and 82.69% aluminum oxide. This detailed breakdown not only helps in understanding the composition but also illustrates how stoichiometry plays a pivotal role in chemical reactions and calculations.