Al(OH)3 + 3HCl = AlCl3 + 3H2O
If 1 tablespoon will be 1 g of mass, then
n(Al(OH)3) = 1g/78g/mol = 0.0128 mol
According to the equation
n(HCl) = 3*n(Al(OH)3) = 3*0.0128 = 0.0384 mol
m(HCl) = 0.0384mol*36.5g/mol = 1.4 g - can be actually neutralized by 1 g tablespoon of Al(OH)3
Comments
Leave a comment