Question

Cobalt forms two common ion oxidation states, Co+3 and Co+3. When solid cobalt is reacted with a gold (III) chloride solution, two reactions are possible, as shown in the following equations:

Reaction 1) AuCl3 (aq)  + Co(s)  → Au(s)  + CoCl3 (aq)

Au+3(aq)  + Co(s)  →   Au(s)  + Co+3(aq)

Reaction 2) 2 AuCl3 (aq)  + 3 Co(s)  →   2 Au(s)  + 3 CoCl2 (aq)

2 Au+3(aq)  + 3 Co(s)  →   2 Au(s)  + 3 Co+2(aq)

If reaction 2 is correct how many grams of solid gold (MW = 196.967) will be produced if 10.000 grams of cobalt metal (MW=58.933) is reacted with an excess of gold (III) chloride? (record your answer to 3 decimal places and do not add a unit ie g)

Answer 1

Reaction 2) 2 AuCl3 (aq) + 3 Co(s) → 2 Au(s) + 3 CoCl2 (aq)

according to balanced euation 3 mol Co produce 2 mole Au . Since
AuCl3 is in excess, we need to consider mols of Co present.

mass of Co = 10 g
molar mass of Co = 196.967 g/mol
number of mols of Co present = 10 g/196.967 g/mol
=0.0507699259 mol

mols of Au produced = 2/3 *mols of Co present
= 2/3 * 0.0507699259 mol
= 0.0338466173 mols

mass of Au produced = Mols of Au formed * molar mass of Au
=0.0338466173 mols* 58.933 g/mol
=1.995 g

Mass of gold = 1.995
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thanks!