A 5.000 g piece of cobalt metal is added to a 106.5 mL of 2.05 M gold(I) nitrate

Question

A 5.000 g piece of cobalt metal is added to a 106.5 mL of 2.05 M gold(I) nitrate. A reaction occurs resulting in the formation of gold metal and cobalt(III) nitrate in solution. a) Write a balanced equation for the reaction (don't forget states of matter) b) If all of the species being oxidized reacts, how much gold metal will form? c) If all of the species being reduced reacts, how much gold metal will form? d) Which is the limiting reagent and what is the theoretical yield of the gold metal? e) Based on the limiting reagent you just determined, how many moles of cobalt(III) nitrate will form and (assuming the volume stays the same), what is the concentration of cobalt(III) nitrate?

Explanation / Answer

a) The balanced reaction is as follows:

Co(s) + 3 AuNO3(aq) -------> Co(NO3)3(aq) + 3 Au (s)

b) In this reaction, cobalt metal oxidises and releases 3 electrons. The half reaction is given as:

Co(s) -------> Co3+(aq) + 3 e-

number of moles of the cobalt (n) = mass of the cobalt (m) / Atomic mass of the cobalt (M)

n = 5.0 g / 58.93 g mol-1

n = 0.0848 mol

From the balanced equation, we know that 3 mol of elemental gold are formed per mol of elemental cobalt

number of moles of the Au = 3 X number of moles of the cobalt

= 3 X 0.0848 mol = 0.2545 mol

mass of Au = number of moles of Au X Atomic mass of Au

= 0.2545 mol X 196.97 g mol-1

mass of Au = 50.13 g

Thus 50.13 g gold will be formed if all the species being oxidised reacts.

c) In this reaction, Au(I) reduces to elemental Au and by accepting one electron. The half reaction is given as:

Au+(aq) + e- -------> Au(s)

number of moles of the Au(I) (n) = concentration of Au(I) salt X Volume of the solution

n = 2.05 mol L-1 X 0.1065 L

n = 0.2183 mol

From the balanced equation, we know that 1 mol of elemental gold is formed per 1 mol of Au(I) salt

number of moles of the Au = number of moles of Au(I) salt = 0.2183 mol

mass of Au = number of moles of Au X Atomic mass of Au

= 0.2183 mol X 196.97 g mol-1

mass of Au = 43.00 g

Thus 43.00 g gold will be formed if all the species being reduced reacts.

d) From subpart b and c, we know that the number of moles of the Au(I) is lesser than the number of moles of the elemental Co and hence Au(I) salt is the limiting reagent. The theoretical yield of elemental Au will be determined by Au(I) salt and is 43.00 g

e) From the balanced equation, we know that 1 mol of Co(III) is formed per 3 mol of Au(I) salt

number of moles of Co(III) salt = number of moles of Au(I) salt / 3

= 0.2183 mol / 3

number of moles of Co(III) salt = 0.0728 mol

concentration of Co(III) salt = number of moles of the Co(III) / Volume of the solution

= 0.0728 mol / 0.1065 L

concentration of Co(III) salt = 0.683 mol L-1

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