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.

explanation:

i think option no d is the correct answer

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