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if the reaction of 5.23×10^38 molecules of gold chloride with excess sodium produces 5.933×10^11 kg of...

if the reaction of 5.23×10^38 molecules of gold chloride with excess sodium produces 5.933×10^11 kg of sodium chloride, what is the percent yield from this reaction? write the balance equation for the reaction as well.

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Answer #1

AuCl3 + 3Na ----> 3NaCl + Au

no of mole of AucL3 = (5.23*10^38)/(6.023*10^23) = 8.68*10^14 mole

no of mole of NaCl(theoretical yield) = 8.68*10^14 *3 = 2.604*10^15 mole

no of mole of NaCl produced (actualyield)= 5.933*10^11/58.5 = 10141880341.9 mole

% of yield = actualyield/theoretical yield*100

           = 10141880341.9/(2.604*10^15)*100

           = 0.0004%

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