According to the equation, 2Al(s) + 6H2O(l) + 2KOH(aq) to make 2K[Al(OH)4](aq) + 3H2(g). How many grams of hydrogen gas would be formed in the reaction of 1.15 grams of Al and excess KOH?

0.06457g of H₂

Explanation:

2Al(s) + 6H₂O(l) + 2KOH(aq) → 2K[Al(OH)₄](aq) + 3H₂(g)

Based on the equation 2 moles of Al produce 3 moles of hydrogen.

First, we need to convert mass of Al to moles and then with the chemical equation find moles and mass of hydrogen:

Moles Al (Molar mass: 26.982g/mol):

1.15g Al * (1mol / 26.982g) = 0.04262 moles Al

Moles H₂:

0.04262 moles Al * (3 moles H₂ / 2 mol Al) = 0.06393 moles H₂

Mass hydrogen (Molar mass: 1.01g/mol):

0.06393 moles H₂ * (1.01g/mol) =

idk what the question is about