Answer to Question #138646 in General Chemistry for Jeanly

Question

The following reaction takes place at high temperatures.
Cr2O3(s) + 2Al(l) → 2Cr(l) + Al2O3(l)
If 50.5 g (Cr2O3) and 12.6 g (Al) are mixed and reacted until one of the
reactants is used up,
a. Which reactant will be the limiting reagent?
b. How many grams of chromium will be formed?
c. How much excess reagent will be left after the reaction?

Answer 1

n (Cr₂O₃) = $m/M$ M (Cr₂O₃) = 152 g/mol

n (Cr₂O₃) = 50,5/ 152 = 0,332 mol

n (Al)=m/M n(Al)=12,6/27=0,47 mol

aluminum is taken in short supply, it is a limiting reagent

n(Al) = n (Cr) n(Cr)=0,47 mol m=0,47*52=24,44g

the chemical reaction is spent n (Cr2O3)=0,235  
It was 0,332  
0,332-0,235=0.097 mol  Cr2O3  left