Gold can be recovered from sea water by re- acting the water with zinc, which is refined from zinc oxide.
2 ZnO(s) + C(s) −→ 2 Zn(s) + CO2(g) 2 Au3+(aq) + 3 Zn(s) −→3 Zn2+(aq) + 2 Au(s) The zinc displaces the gold in the water. What mass of gold can be recovered if 9.7 g of ZnO and an excess of sea water are available?
Answer in units of g
mols ZnO = grams ZnO/molar mass ZnO
Using the coefficients in the balanced equation, convert mols ZnO to mols Au.
Now convert mols Au to grams. g = mols x atomic mass.
To find the mass of gold that can be recovered, we need to determine the limiting reactant between zinc oxide (ZnO) and gold ions (Au3+).
To do this, we need to calculate the moles of ZnO available in 9.7 g of ZnO. First, we find the molar mass of ZnO, which is 65.38 g/mol (the atomic mass of zinc is 65.38 g/mol and oxygen is 16.00 g/mol).
Moles of ZnO = mass of ZnO / molar mass of ZnO
= 9.7 g / 65.38 g/mol
≈ 0.1483 mol
Next, we use the stoichiometry of the reaction to determine the moles of gold (Au) that can be recovered. From the balanced equation, we can see that the ratio of moles of ZnO to moles of Au is 2:2. Therefore, the moles of Au will be equal to the moles of ZnO.
Moles of Au = Moles of ZnO
= 0.1483 mol
Finally, we calculate the mass of gold (Au) using the molar mass of gold, which is 196.97 g/mol.
Mass of gold = Moles of Au x Molar mass of Au
= 0.1483 mol x 196.97 g/mol
≈ 29.2 g
Therefore, approximately 29.2 grams of gold can be recovered if 9.7 grams of ZnO and an excess of seawater are available.