How many grams of alcl3 are produced when 3 grams of al (oh) 3 react completely with excess hcl?

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