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?

The theoretical yield of chromium will be 15.05 g

Solution:

The balance chemical equation

2Al + Cr2o3 - -> Al2o3 + 2Cr

From given mass, we can calculate moles

mass of Al = 8 g mass of Cr2O3 = 40g

Molar mass of Al = 26.98g/mol Molar mass of Cr2O3 = 151.9g/mol

Mole of Al = 8/26.98 Mole of Cr2O3 = 40/151.9

Mole of Al = 0.29 mole Mole of Cr2O3 = 0.26 mol

Here, we have to find out the limiting reactant (The reactant which will give the least amount of product) so from balance chemical equations

No. of mole of Al = No. of moles of Cr

2 = 2

0.29 = 2/2 * 0.29 = 0.29 mole

No. of mole of Cr2O3 = No. of moles of Cr

1 = 2

0.26 = 2/1 * 0.26 = 0.52 mole

So we can say, Al is a limiting reactant so

No. of mole of Al = No. of moles of Cr

2 = 2

0.29 = 2/2 * 0.29 = 0.29 mole

Moles of Cr = 0.29 mol

Molar mass of Cr = 51.9 g/mol

mass of Cr = 0.29 * 51.9 = 15.05 g

Conclusion:

The calculated mass from balance chemical equation of Cr will be 15.05 g.