45.7 grams of calcium chloride reacts with an excess of aluminum oxide. how many grams of aluminum chloride will be produced

Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol

                                               3CaCl2+Al2O3 >3CaO +2AlCl3
mole from reaction              3 mol                                              2 mol
mass from reaction         3mol* 111.1g/mol                             2 mol*133.5g/mol
                                               333.3 g                                            267.0 g
mass from problem              45.7 g                                               x g

Proportion:
  333.3 g  CaCl2     267.0 g AlCl3
  45.7 g   CaCl2      x g    AlCl3

x=45.7*267.0/333.3= 36.6 g AlCl3

Its strengthits conductivity of heat and electricity its shiny colourmalleable and ductility