Question

What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum according to the chemical equation below?
2Al + Cr2O3 --> Al2O3 + 2Cr
A)7.7 g
B)15.4 g
C) 27.3 g
D) 30.8 g
E)49.9 g

Answer 1

Answer:

B) 15.4 g

Explanation:

First, we need to know which of the reagents is limiting. Let's do the stoichiometry calculus for Al and test if it is limiting.

For 2 moles of Al it's necessary 1 mol of Cr2O3. The molar masses of the elements are:

Al = 27 g/mol; O = 16 g/mol; Cr = 52 g/mol

So, the molar masses of the compounds are:

Cr2O3 = 2x52 + 3x16 = 152 g/mol

Al2O3 = 2x27 + 3x16 = 102 g/mol

The stoichiometry calculus is:

2 moles of Al ---------------- 1 mol of Cr2O3

2x27 g of Al ----------------- 152 g of Cr2O3

8 g of Al ----------------------- x g of Cr2O3

By a direct simple three rule:

54x = 1216

x = 22.52 g

So, 22.52 g of Cr2O3 must react with 8.00g of Al, then Cr2O3 is in excess, and Al is the limiting reagent.

Now, doing the stoichiometry calculus between the limiting reagent and chromium:

2 moles of Al ----------------------- 2 moles of Cr

1 mol of Al ---------------------------- 1 mol of Cr

27 g of Al ---------------------------- 52 g of Cr

8 g of Al ------------------------------ y

By a direct simple three rule:

27y = 416

y = 15.41 g