how many moles will be left after the reaction is complete?

[Smooth Operater]

Given 1 mol of Ag and 0.4 mol of S, which is the excess reagent and how many moles will be left after the reaction is complete?

2Ag + S --> Ag2S

I know Ag is the excess b/c 1 mol of Ag yield 1/2 mol of Ag2S. And 0.4 mol of S yield 0.4 mol of Ag2S.

But how can I find how many moles will be left after the reaction is complete? Thanks!

---

Members don't see this ad.

 

[fancymylotus]

1 mol of Ag *1 mol of S/2 mol Ag= 1/2 or 0.5
0.4 mol of S * 2 mol of Ag/1 mol of S=0.8
so Ag is limiting reactant.
so 1/2mol of Ag*1 mol of Ag2S/2 mol of Ag* 246gAg2S/1mol Ag2S= answer

i HATE dimensional analysis i always get these questions wrong

 

[Manyak222]

I think i visualized it wrong lol. ill get back to this one later

But how can I find how many moles will be left after the reaction is complete? Thanks![/QUOTE]

 

[Manyak222]

I think i visualized it wrong lol. ill get back to this one later

ok here goes. since you need twice the Ag, its really like you have only .5 moles of it compared to S. thus you have .1 moles of Ag left over

But how can I find how many moles will be left after the reaction is complete? Thanks!

[/QUOTE]

 

[Manyak222]

ya but by how much?I think i had it right the first time, cuz .4 moles of s will react with .8 moles of ag. so there should be .2 moles of Ag. Thats my FINAL answer

sorry did that wrong earlier but
1 mol of Ag/2=.5
and 0.4 mol of S/1=.4
so Ag is in excess

 

[Notoriousjae]

Given 1 mol of Ag and 0.4 mol of S, which is the excess reagent and how many moles will be left after the reaction is complete?

2Ag + S --> Ag2S

I know Ag is the excess b/c 1 mol of Ag yield 1/2 mol of Ag2S. And 0.4 mol of S yield 0.4 mol of Ag2S.

But how can I find how many moles will be left after the reaction is complete? Thanks!

sorry did that wrong earlier because my brain is fried from too much DAT but
1 mol of Ag/2=.5
and 0.4 mol of S/1=.4
so Ag is in excess and S is limiting.
so you find the S which is limiting in grams and then minus it from the Ag2S in grams to get whats left over in excess and convert that to moles to find out how many moles in excess.
am i right? i cant really think straight right now.

 

[NilamPatel]

since S is the limiting reactant you use that to find out how many moles you get:



moles of Ag used (use the mole ratio given from equation):
.4 moles of S * 2 moles of Ag/1 mole of S = .8 moles of Ag used

moles of Ag available : 1 mole available

excess:
available - used = 1-.8 = .2 moles in excess