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?

Answer 1

Answer:

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.

Answer 2

Chromium is an element with atomic number 24. The 40 gram of chromium oxide reacts with 8 gram of aluminium to form aluminium oxide and chromium. The mass of chromium will be 15.05 gram.

The given equation can be represented as:

$\text {2 Al} + \text {Cr}_{2} \text O_{3}\; \rightarrow\; \text {Al}_{2}\text O_{3} + \text { 2 Cr}$

Given that:

• Mass of Aluminum = 8 gram
• Molar Mass of Aluminum = 26.9 g/mol
• Moles of Aluminum = $\frac{8}{26.9}& =0.29\; \text {moles}$

Similarly, for Chromium Oxide:

• Mass of Chromium Oxide= 40 gram
• Mass of Chromium Oxide= 151.9 g/mol
• Moles of Mass of Chromium Oxide=$\frac{40}{151.9}& =0.26\; \text {moles}$

Now, the limiting reactant can be found out as:

Number of moles of Al = Number of moles of Cr

                                  2 = 2

                            0.29 = 0.29 moles

Number of moles of Chromium oxide  = Number of moles of Chromium

                                                          1    =     2

                                                    0.26   =  2 X 0.26  = 0.52  moles

Thus, aluminum is a limiting reactant.

Calculating the mass of chromium, we get:

• Moles of Cr = 0.29
• Molar Mass of Cr = 51.9 gram/mol
• Mass of Cr = m = n X M = 15.05 grams

Therefore, the mass of Cr requires in the equation will be 15.05 gram.

To know more about molar mass, refer to the following link:

https://brainly.com/question/14852862?referrer=searchResults