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A chemist wants to extract the gold from 54.9 g of AuCl3*2H20 *gold(III) chloride dyhydrate) by electrolysis of an aqueous solution. What mass of gold could be obtained from this sample? Answer in units of g

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From the information provided in the question, the mass of gold that can be obtained from this sample is 31.5 g of Au.

Given that the reduction of gold III occurs thus;

Au^3+(aq) + 3e ------> Au(s)

Number of moles of gold(III) chloride dihydrate = mass/molar mass = 54.9 g /339.35 g/mol = 0.16 moles

Number of moles of Au^3+ in  AuCl3.2H20 =  0.16 moles

Now;

1 mole of Au has a mass of 197 g/mol

0.16 moles of Au has a mass of 0.16 moles ×  197 g/mol/1 mole

= 31.5 g of Au.

The mass of gold that can be obtained from this sample is 31.5 g of Au.

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