Find the mass of AlCl3 that is produced when 7.7 grams of Al2O3 react with 13.2 g of HCl according to the following equation. (Molar masses: molar mass Al2O3 = 101.96 g/mol, molar mass HCl = 36.45 g/mol, molar mass AlCl3 = 133.34 g/mol, molar mass H2O = 18.0 g/mol) Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(aq)
Answers
Answer:
Al2O3 + 6HCl ==> 2AlCl3 + 3H2O ... balanced equation
moles Al2O3 present = 30.0 g x 1 mol/101.96 g = 0.294 moles
moles HCl present = 30 g HCl x 1 mol/36.5 g = 0.822 moles HCl
HCl is LIMITING as it takes 6 moles HCl for each 1 mol Al2O3 and here that is not enough. It will run out first.
(note: an easy way to determine limiting reactant is to divide moles of each reactant by the corresponding coefficient in the balanced equation, and the smaller value represents the limiting reactant)
Now, using the limiting reactant, we find the moles and mass of AlCl3 that can be formed.
0.822 moles HCl x 2 moles AlCl3/6 moles HCl = 0.274 moles AlCl3 formed
mass of AlCl3 = 0.274 moles AlCl3 x 133 g/mole = 36.4 g AlCl3 formed
MARK ME AS BRAINLIEST.