[gary006]

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 4.00 g of ZnO and an excess of sea water are available?

2 ZnO(s) + C(s) -> 2 Zn(s) + CO2(g)
2 Au3+(aq) + 3 Zn(s) -> 3 Zn2+(aq) + 2 Ag(s)


I converted 4.00 g of ZnO to .049146 mol and then I multiplied that by the mass of Au and got 9.68 buts its wrong.

Can anyone help? thanks in advance.

[kkrizka]

You found the number of moles  of Zn right, but forgot a step afterwards. The ratio of moles of Zn consumed to moles of Au produced is not 1:1 but 3:2.

Hope that helps.

[gary006]

Ok so

.049146 mol Zn * (3 mol Au/2 mol Zn) = .073719 mol Au * mass of Au = 14.5202 is that right?

[sdekivit]

2 ZnO(s) + C(s) -> 2 Zn(s) + CO2(g)
2 Au3+(aq) + 3 Zn(s) -> 3 Zn2+(aq) + 2 Au(s)



the second reaction is wrong   --> needs to be Au(s)

but apart of that: yyou should multiply number of mols Zn with 2/3

[gary006]

2 ZnO(s) + C(s) -> 2 Zn(s) + CO2(g)
2 Au3+(aq) + 3 Zn(s) -> 3 Zn2+(aq) + 2 Au(s)



the second reaction is wrong   --> needs to be Au(s)

but apart of that: yyou should multiply number of mols Zn with 2/3

ok thanks so what do I need to do after multypying the number of mols Zn with 2/3? I only have 1 more chance to get it right.

[plu]

ok thanks so what do I need to do after multypying the number of mols Zn with 2/3? I only have 1 more chance to get it right.

You've got to keep a good head on your conversions there, mate.  You started with .049146 mol of ZnO.  From the first equation, you know that the ratio ZnO:Zn is 2:2 (1:1).  Therefore, you've got .049146 mol of Zn.  Now, from the second equation, you know that the ratio Zn:Au is 3:2, as kkrizka pointed out.  Thus, you can convert moles of Zn to moles of Au by multiplying by 2/3 (as you have done).  Then convert moles of Au into grams of Au by multiplying by the molar mass of Au to find the mass of gold that can be recovered  

[gary006]

You've got to keep a good head on your conversions there, mate.  You started with .049146 mol of ZnO.  From the first equation, you know that the ratio ZnO:Zn is 2:2 (1:1).  Therefore, you've got .049146 mol of Zn.  Now, from the second equation, you know that the ratio Zn:Au is 3:2, as kkrizka pointed out.  Thus, you can convert moles of Zn to moles of Au by multiplying by 2/3 (as you have done).  Then convert moles of Au into grams of Au by multiplying by the molar mass of Au to find the mass of gold that can be recovered  

Thanks but the answer I get is 6.4531 g right? But that answer is wrong, so what could I be doing wrong?

[kkrizka]

I get 6.454519392, but to proper significant figures it should be 6.45 so we got the same answer. Could you tell us the correct answer?

[sdekivit]

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 4.00 g of ZnO and an excess of sea water are available?

2 ZnO(s) + C(s) -> 2 Zn(s) + CO2(g)
2 Au3+(aq) + 3 Zn(s) -> 3 Zn2+(aq) + 2 Au(s)

4,00 g ZnO = 0,049152 mol ZnO

--> this yields the same amount of Zn

then the amount of mols Au that can be produced:

0,049152 * (2/3) = 0,032681 mol Au

--> that's 6,46 g Au (s)