Using the balances eqution and assuming all other reagents are in excess, how many hydrogen would be produced when 1.00 g of aluminum react?

What balanced equation?

Balance the equation

acid + aluminum> H2 gas + aluminum salt.

Lets look at some examples

6HCl + 2Al>>3H2 + 2AlCl3
so for each two Aluminum moles, you get 3 moles of hydrogen gas.

3H2SO4 + 2Al>> Al2(SO4)3 + 3H2
Again, for each two aluminum moles reacted, one gets three hydrogen moles.

Try it for other acids. So find the moles of aluminum in 1 gram, multiply it by 3/2, that is the moles of H2 you will get.

To determine the number of hydrogen atoms produced when 1.00 g of aluminum reacts, we need to write a balanced chemical equation for the reaction.

The balanced equation for the reaction between aluminum and hydrochloric acid can be written as:

2Al + 6HCl → 2AlCl3 + 3H2

From this equation, we can see that for every 2 moles of aluminum (Al), 3 moles of hydrogen gas (H2) are produced.

To find the number of moles of aluminum in 1.00 g, we need to use the molar mass of aluminum. The molar mass of aluminum is 26.98 g/mol.

Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum
= 1.00 g / 26.98 g/mol = 0.037 mol

Since the stoichiometry of the reaction is 2 moles of aluminum to 3 moles of hydrogen gas, we can use the mole ratio to find the number of moles of hydrogen:

Number of moles of hydrogen gas = Number of moles of aluminum × (3 moles of H2 / 2 moles of Al)
= 0.037 mol × (3/2) = 0.0555 mol

Finally, to find the number of hydrogen atoms, we need to multiply the number of moles of hydrogen gas by Avogadro's number (6.022 x 10^23):

Number of hydrogen atoms = Number of moles of hydrogen gas × Avogadro's number
= 0.0555 mol × 6.022 x 10^23 = 3.34 x 10^22 hydrogen atoms

Therefore, when 1.00 g of aluminum reacts, approximately 3.34 x 10^22 hydrogen atoms are produced.