Question

Chromium (II) and Oxygen (diatomic molecule) combine to form chromium (II) oxide. Write, name and balance the reactants and products and answer the following questions. 

How many moles of oxygen react with 4.50 moles of Chromium (III)? 

How many grams of chromium (III) oxide are produced when 24.8 g of Chromium (III) reacts? 

When 26.0 grams of Chromium (III) reacts with 8.00 g of O2, how many grams of Chromium (III) and Oxygen can form?

Answer 1

1) The ballanced equation is

4 Cr+3 + 3O2 -----------------> 2 Cr2O3

2) a)

From the stoichiometric equation

4 moles of Cr+3 reacts with 3 moles of Oxygen.

To react with 4.5 moles of Cr oxygen required = 4.5x 3 /4 = 3.375 mol

b)

4 moles of Cr+3 reacts with 3 moles of Oxygen to give 2 moles of Cr2O3

4 x 52 g of Cr+3 reacts wth 3 x 32 g of oxygen to give 2 x 152 g of Cr2O3

If 24.8g of Cr+3 is used mass of Cr2O3 = 24.8 x 2 x 152 / 4x52

= 36.246 g

c)

4 Cr+3 + 3O2 -----------------> 2 Cr2O3

26g/52 8/32 0 initial moles

= 0.5 =0.25 0

To know the limiting reagent

0.5/4 = 0.125 0.25/3 =0.0833

Thus the ratio of O2 is less, and hence o2 is the limiting reagent .

Thus

3x32 g of O2 give 2 x 152 g of CR2O3

8 g of O2 can give = 8 x 2 x 152 /3x32

= 25.33 g of Cr2O3