Gold can be recovered from sea water by reacting the water with zinc, which is refined from zinc oxide. The zinc displaces the gold in the water. What mass of gold can be recovered if 2.00g of ZnO and an excess of sea water are available?
2ZnO + C -> 2Zn + CO2
2Au^+3 + 3Zn -> 3Zn^+2 + 2Au
6ZnO+ 3C + 4Au+3 >> 6Zn+2 + 4Au + 3CO2
so for each mole of ZnO consumed, one gets 4/6 mole of Gold
massgold=2gZnO*1moleZnO/molmassAuO * 2 moleAu/3moleZnO * molmassAu/1moleAu
Could you break that down a little? I don't completely understand.
You're just working another stoichiometry problem.
2 g ZnO = how many moles.
moles = grams/molar mass.
??moles ZnO x (4 moles Au/6 moles ZnO = xx moles Au.
Convert moles Au to grams.
To find the mass of gold that can be recovered, we need to use stoichiometry and convert the given mass of ZnO to the mass of gold.
First, we need to determine the molar mass of ZnO. The molar mass of Zn is 65.38 g/mol and the molar mass of O is 16.00 g/mol. Therefore, the molar mass of ZnO is:
Molar mass of ZnO = (65.38 g/mol) + (16.00 g/mol) = 81.38 g/mol
Next, we need to use the balanced chemical equation to determine the stoichiometric ratio between ZnO and Au:
2ZnO + C -> 2Zn + CO2
2Au^+3 + 3Zn -> 3Zn^+2 + 2Au
From the equations, we can see that 2 moles of ZnO are required to produce 2 moles of Au. Therefore, the stoichiometric ratio is:
2 moles of ZnO : 2 moles of Au
Now, we can calculate the moles of ZnO available:
Moles of ZnO = Mass of ZnO / Molar mass of ZnO
Moles of ZnO = 2.00 g / 81.38 g/mol ≈ 0.0246 mol
Since there is an excess of sea water (which contains Au^+3), all the ZnO will be used to react with the gold ions. Therefore, the moles of gold that can be recovered will be the same as moles of ZnO.
Moles of Au = Moles of ZnO = 0.0246 mol
Finally, we can calculate the mass of gold using the molar mass of gold (197.0 g/mol):
Mass of gold = Moles of Au × Molar mass of Au
Mass of gold = 0.0246 mol × 197.0 g/mol ≈ 4.84 g
Therefore, approximately 4.84 grams of gold can be recovered.