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

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

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120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890