The Volume of Hydrogen Gas Produced When Aluminum Reacts with Hydrochloric Acid at STP

When aluminum (Al) reacts with hydrochloric acid (HCl), a reaction takes place that is commonly used in qualitative analysis to produce hydrogen gas (H?). The balanced chemical equation for this reaction is as follows:

Balanced Equation:

2Al (s) 6HCl (aq) → 2AlCl? (aq) 3H? (g)

This equation shows that for every 2 moles of aluminum that react, 3 moles of hydrogen gas are produced. In this article, we will walk through the detailed steps to determine the volume of hydrogen gas (H?) produced when 6.80 grams of aluminum reacts under standard temperature and pressure (STP).

Molar Mass Calculation

The molar mass of aluminum (Al) is 26.98 grams per mole.

Using the given mass of aluminum (6.80 grams) and its molar mass, we can find the number of moles of aluminum:

[math] n_{Al} frac{6.80text{g}}{26.98text{g/mol}} 0.252text{mol} ]

Calculation of Hydrogen Gas Produced

According to the balanced equation, 1 mole of aluminum produces 1.5 moles of hydrogen gas. Therefore, 0.252 moles of aluminum will produce:

[math] 0.252text{mol} times 1.5 0.378text{mol of H}_{2} ]

Volume of Hydrogen Gas at STP

At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 liters. Using this, we can find the volume of 0.378 moles of hydrogen gas:

[math] 0.378text{mol} times 22.4text{L/mol} 8.47text{L} ]

Thus, the volume of hydrogen gas produced when 6.80 grams of aluminum reacts with hydrochloric acid at STP is 8.47 liters.

Reaction Representation:

We can represent the oxidation of aluminum metal with hydrochloric acid by the following equation:

[math] text{Al} (s) 3 text{HCl} (aq) rightarrow text{AlCl}_{3} (aq) frac{3}{2} text{H}_{2} (g) ]

By applying the molar mass calculation to the given 6.80 grams of aluminum:

[math] n_{text{aluminum}} frac{6.80text{g}}{26.98text{g/mol}} 0.252text{mol} ]

Therefore, the reaction will produce:

[math] frac{3}{2} equiv text{dihydrogen gas i.e. 0.756 mol} ]

The volume at STP for this amount of hydrogen gas is:

[math] 0.756text{mol} times 22.4text{L/mol} 16.88text{L} ]

Conclusion:

By following the steps outlined above, we can easily determine the volume of hydrogen gas produced in a reaction involving aluminum and hydrochloric acid at STP. This process not only provides insight into the stoichiometry of the reaction but also helps in understanding the principles of gas volume calculations at standard temperature and pressure.