How much hydrogen gas (H2) forms when 54 g of aluminum reacts with excess HCI ?
To determine the amount of hydrogen gas (H2) formed when aluminum (Al) reacts with excess hydrochloric acid (HCl), we need to use the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between aluminum and hydrochloric acid is:
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
From the equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.
To calculate the amount of hydrogen gas generated, we need to follow these steps:
1. Convert the given mass of aluminum (54 g) into moles.
To do this, divide the mass of aluminum by its molar mass. The molar mass of aluminum (Al) is 26.98 g/mol.
Moles of Al = Mass of Al / Molar mass of Al
= 54 g / 26.98 g/mol
= 2 moles of Al
2. Use the stoichiometric ratio between aluminum and hydrogen from the balanced equation.
From the balanced equation, we know that 2 moles of Al react to produce 3 moles of H2.
Moles of H2 = (Moles of Al) x (Moles of H2 / Moles of Al)
= 2 moles of Al x (3 moles of H2 / 2 moles of Al)
= 3 moles of H2
3. Convert moles of hydrogen gas to grams.
To do this, multiply the number of moles by the molar mass of hydrogen (H2), which is 2.02 g/mol.
Mass of H2 = Moles of H2 x Molar mass of H2
= 3 moles of H2 x 2.02 g/mol
= 6.06 g
Therefore, 54 g of aluminum will react with excess hydrochloric acid to form 6.06 g of hydrogen gas (H2).
first we write the chemical reaction:
Al + HCl -> AlCl3 + H2
then we balance it:
2Al + 6HCl -> 2AlCl3 + 3H2
also we need to look for the masses of Al. from the periodic table,
Al = 27 g/mol
therefore,
54 g Al * (1 mol Al / 27 g Al) * (3 mol H2 / 2 mol Al) = 3 mol H2
or in grams, (atomic mass of H = 1 g/mol)
3 mol H2 * (2 g H2 / 1 mol H2) = 6 g H2
hope this helps~ :)