Question

The reaction of Cr_2O_3 with silicon metal at high temperatures will make chromium metal. 2Cr_2O_3 (s) + 3Si (s) rightarrow 4Cr (l) + 3SiO_2 (s) The reaction is begun with 182 g of Si and 1.40 times 10^2 g of Cr_2O_3. Which reactant is the limiting reactant? How many grams of the excess reactant are left after the reaction is complete?

Answer 1

number of moles of Si = 182 g / 28.0855 g.mol^-1 = 6.48 mole

number of moles of Cr2O3 = 1.40*10^2 g / 151.99 g.mol^-1 = 0.921mole

from the balanced equation we can say that

2 mole of Cr2O3 requires 3 mole of Si

so 0.921 mole of Cr2O3 will require

= 0.921 mole of Cr2O3*(3 mole of Si / 2 mole of Cr2O3)

= 1.38 mole of Si

but we have 6.48 mole of Si so Si is the excess reactant. Cr2O3 is limiting reactant

Number of moles of excess reactant = 6.48 mole - 1.38 mole of Cr2O3 = 5.1 mole of Si left

1 mole of Si = 28.0855 g.mol^-1

5.1 mole of Si = 143.2 g of excess reactant left