Question

Chromium and oxygen combine to form chromium (III) oxide. 4Cr(s)+3O2(g)→2Cr2O3(s)

How many moles of O2 react with 4.85 mol of Cr?

How many grams of Cr2O3 are produced when 23.6 g of Cr reacts?

When 25.0 g of Cr reacts with 7.20 g of O2, how many grams of Cr2O3 can form?

If 72.1 g of Cr and 68.0 g of O2 are mixed, and 87.0 g of Cr2O3 is actually obtained, what is the percent yield of Cr2O3 for the reaction?

Express your answer with the appropriate units.

Answer 1

Chromium and oxygen combine to form chromium (III) oxide. 4Cr(s)+3O₂(g)→2Cr₂O₃(s)

As per this balanced equation 4 moles of chromium combines with 3 moles of oxygen gas to form 2 moles of chromium (III) oxide

1. How many moles of O₂ react with 4.85 mol of Cr?

If 4 moles of chromium combines with 3 moles of oxygen then ¾ x 4.85 moles of oxygen will combine with 4.85 moles of chromium = 3.63 moles of oxygen.

2. How many grams of Cr₂O₃ are produced when 23.6 g of Cr reacts?

MW of Chromium is 51.996 g /mol so 23.6 g of chromium will be

23.6/51.996 = 0.4539 moles

Mw of Cr₂O₃ is 151.99 g/mol

Since 4 moles of chromium produces 2 moles of Cr₂O₃

0.4539 moles of chromium will produce 0.4539/2 = 0.227 moles of Cr₂O₃

Which will be 0.227 moles x 151.99 g/mol = 34.47 g of Cr₂O₃

3. When 25.0 g of Cr reacts with 7.20 g of O2, how many grams of Cr2O3 can form?

25.0 g of chromium will be 25/51.996 = 0.4808 moles of chromium

7.2 g of O₂ will be (MW of oxygen is 16g/mol) 7.2/32 = 0.225 moles of oxygen

So the limiting reagent in this case is oxygen

3 moles of oxygen give 2 moles of Cr₂O₃

so 0.225 moles of oxygen will give 0.225 x 2/3 = 0.15 moles of Cr₂O₃

0.15 moles of Cr₂O₃ is 0.15 mol x 151.9g/mol = 22.78 g of Cr₂O₃

4) If 72.1 g of Cr and 68.0 g of O2 are mixed, and 87.0 g of Cr2O3 is actually obtained, what is the percent yield of Cr2O3 for the reaction?

72.1 g of Cr will be 72.1/51.996 = 1.38 moles

68 g of oxygen will be 68/32 = 2.125 moles of oxygen

So the limiting reagent is chromium

Now 4 moles of chromium give 2 moles of Cr₂O₃

1.38 moles of Cr will give 1.38 x 2/4 = 0.69 moles of Cr₂O₃

0.69 moles of Cr₂O₃ is 0.69 mol x 151.9 g/mol = 104.81 g of Cr₂O₃

So for 100% conversion we should have got 104.81 g of Cr₂O₃

Since we have got 87.0 g the percentage yield will be

87.0/104.81 x 100 = 83%