Question

What is the percent yield of NaCN in the reaction, 4NaAu(CN)2 + 4NaOH arrow 4Au + 8NaCN + O2 + 2H2O, if 3.00 grams of NaCN are actually produced when 10.0 grams of NaAu(CN)2 are decomposed? (This is the process of the extraction of gold from its ore.)

Can anyone walk me through this?

Answer 1

1. convert 10.0 grams of NaAu(CN)2 to moles by dividing by the molar mass
2. Using the equation, we can see that for every 4 moles of NaAu(CN)2 used, 8 moles of NaCN are produced. We can multiply the number of moles of NaAu(CN)2 by 2 (since 8/4 = 2) to find the number of moles of NaCN produced.
3. convert moles of NaCN to grams by multiplying by the molar mass
4. we have the number of grams of NaCN that we would predict (the theoretical yield). we find the theoretical yield by calculating

percent yield = 100*(actual yield)/theoretical yield

in this case, the actual yield is the number of grams of NaCN produced (3.00 grams) and the theoretical yield is the number of grams predicted (obtained in step 3)

Answer 2

Step 1 - 4 NaAu(CN)2 + 4 NaOH -> 4 Au + 8 NaCN + O2 + 2 H2O

Step 2 - (10.0 g NaAu(CN)2) / (271.99134 g NaAu(CN)2/mol) x (8 mol NaCN / 4 mol NaAu(CN)2) x (49.00727 g NaCN/mol) = 3.60 g NaCN in theory

Step 3 - (3.00 g NaCN) / (3.60 g) = 0.833 = 83.3% yield

This is what I said for my answer, am I right?

Answer 3

yup, everything looks good to me