the grams of Alcl3 that are produced when 3 grams of Al(OH)3 react completely with excess Hcl is calculated as follows
write the reacting equation
Al(Oh)3 + 3HCl = AlCl3 + 3H20
find moles of Al (Oh)3 used
moles= mass /molar mass
= 3g/ 78 g/mol= 0.0385 moles
by use of mole ratio between Al(OH)3 to AlCl3 which is 1:1 this implies that the moles of AlCl3 is = 0.0385 moles
mass of Alcl3 = moles x molar mass
= 0.0385 mol x133 .5 g/mol = 5.14 grams
