What is the maximum mass of H2 formed, when 0.380 g of aluminum are reacted with 55.0 mL of 8.00 M HCl? The other product is AlCl3
To find the maximum mass of H2 formed during the reaction, we need to determine the limiting reactant. The limiting reactant is the reactant that gets completely consumed and limits the amount of product produced.
1. Start by calculating the number of moles of aluminum (Al) using its molar mass. The molar mass of aluminum is 26.98 g/mol.
Number of moles of Al = 0.380 g Al / 26.98 g/mol Al
2. Next, calculate the number of moles of hydrogen chloride (HCl) using its concentration and volume. The given concentration is 8.00 M, and the given volume is 55.0 mL.
Number of moles of HCl = (8.00 mol/L) * (0.0550 L)
3. Use the balanced chemical equation to find the stoichiometric ratio between Al and H2. From the equation: 2Al + 6HCl --> 2AlCl3 + 3H2
The stoichiometric ratio between Al and H2 is 2:3 (2 moles of Al produce 3 moles of H2).
4. Compare the moles of Al and HCl. The reactant that is present in a smaller amount is the limiting reactant.
Let's compare the moles of Al and HCl using the stoichiometric ratio:
Moles of H2 (from Al) = (2 moles of Al / 2 moles of Al) * (3 moles of H2)
Moles of H2 (from HCl) = (6 moles of HCl / 2 moles of Al) * (3 moles of H2)
Compare the two values and identify the limiting reactant.
5. Once you have identified the limiting reactant, calculate the maximum mass of H2 produced using its molar mass. The molar mass of hydrogen (H2) is 2.02 g/mol.
Maximum mass of H2 = Moles of H2 * Molar mass of H2
By following these steps, you can find the maximum mass of H2 formed during the reaction.