Determining the Mass of Aluminum Required for a Reaction with Hydrochloric Acid

Understanding the relationship between reactants and products in chemical reactions is a fundamental concept in chemistry. One common example of such a reaction involves aluminum and hydrochloric acid. Let's explore how to determine the mass of aluminum, in grams, required to completely react with a given volume and concentration of hydrochloric acid.

Chemical Reaction and Stoichiometry

The balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl) is:

2 Al(s) 6 HCl(aq) → 2 AlCl3(aq) 3 H2(g)

Key components in this equation are the stoichiometric coefficients, which show the molecular ratios in the reaction. Given this information, we can determine how much aluminum is required to react with a certain amount of hydrochloric acid.

Step-by-Step Calculation

Given Data

Volume of HCl: 1.65 L Concentration of HCl: 0.350 M (Molarity)

Step 1: Calculate Moles of HCl

The number of moles (n) of a substance can be calculated using the formula:

n Molarity (M) × Volume (V) in liters

Application:

$$ntext{ (HCl)} 0.350 text{ M} times 1.65 text{ L} 0.5775 text{ moles}$$

Step 2: Determine Moles of Aluminum Required

From the balanced equation, we know that 6 moles of HCl react completely with 2 moles of Al. Therefore, the mole ratio of HCl to Al is 6:2 or 3:1.

Hence, the moles of Al required can be calculated as:

$$text{Moles of Al} frac{2}{6} times text{Moles of HCl} frac{1}{3} times 0.5775 text{ moles} 0.1925 text{ moles}$$

Step 3: Calculate the Mass of Aluminum Required

The molar mass of aluminum (Al) is approximately 26.98 g/mol. The mass of Al required can be calculated using the formula:

$$text{Mass of Al} text{Moles of Al} times text{Molar Mass of Al}$$

Substituting the values:

$$text{Mass of Al} 0.1925 text{ moles} times 26.98 text{ g/mol} approx 5.194 text{ g}$$

With three significant figures, the mass of aluminum required is 5.19 g.

Conclusion

In summary, the mass of aluminum required to completely react with 1.65 L of 0.350 M HCl can be calculated by first determining the moles of HCl, then using the stoichiometric ratio to find the moles of Al, and finally converting moles of Al to grams using its molar mass. This calculation demonstrates the principles of stoichiometry and the importance of accurate chemical formulas and balanced equations in determining reaction outcomes.

Chemical Reaction Summary

2 Al(s) 6 HCl(aq) → 2 AlCl3(aq) 3 H2(g)

The reaction produces aluminum chloride (AlCl3) and hydrogen gas (H2), with the balanced equation showing that 6 moles of HCl react with 2 moles of Al, producing 2 moles of AlCl3 and 3 moles of H2.

Additional Insights

Understanding such stoichiometric calculations is crucial in not only laboratory settings but also in practical applications where precise chemical reactions are necessary. Whether it's in manufacturing processes, environmental chemistry, or pharmaceuticals, accurate determination of reactant amounts is essential for successful outcomes.

Key Takeaways

Use the balanced chemical equation to establish the mole ratio between reactants. Calculate the moles of each substance using the given volume and concentration. Convert the moles of the required substance to grams using its molar mass.

By following these steps and understanding the underlying principles of stoichiometry, one can accurately determine the necessary amounts of reactants for various chemical reactions.