Question

Cobalt III hydroxide and nitric acid react according to the following balanced equation:

Co(OH)3 (aq) + 3 HNO3 (aq) --> Co(NO3)3(aq) + 3 H2O(l)

a. What mass of cobalt III hydroxide is needed to make 5.2 liters of a 0.42 M cobalt III hydroxide solution?

b. How many milliliters of 1.6 M nitric acid is needed to completely react with the amount of cobalt III hydroxide in Part 2a above?

c. Based on Parts 2a and 2b above, how many moles of water would be produced?

Answer 1

The mass of cobalt (III) needed is
m = 5.2 L (0.42 mol/L) ( 93 g/mol)
m = 97.65 g

The volume of nitric acid needed is
V = 5.2 L (0.42 mol/L) (3 mol / 1 mol) (1000 mL/1.6 mol)
V = 1968.75 mL

The moles of water produced is
n = 5.2 L (0.42 mol/L) (3 mol / 1 mol)
n = 3.15 moles