In the reaction of aluminum with sulfuric acid, 3.42 g of aluminum sulfate was formed.

In the reaction of aluminum with sulfuric acid, 3.42 g of aluminum sulfate was formed. Determine the mass and amount (mol) of aluminum reacted.

Given:
m (Al2 (SO4) 3) = 3.42 g
To find:
m (Al)
n (Al)
Decision:
3H2SO4 + 2Al = Al2 (SO4) 3 + 3H2
n (Al2 (SO4) 3) = m / M = 3.42 g / 342 g / mol = 0.01 mol
n (Al2 (SO4) 3): n (Al) = 1: 2
n (Al) = 0.01 mol * 2 = 0.02 mol
m (Al) = n * M = 0.02 mol * 27 g / mol = 0.54 g
Answer: 0.01 mol, 0.54 g