How to solve the 31st one
I tried to find the molar concentration of ions but 2 moles of Ag2SO4 are left in the reaction I don’t understand what volume to take if some volume is already involved in reaction can someone help
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@Dexter
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2L of 2M Ag2SO4 means amount of moles = 4 (since moles/volume = molarity)
4L of 1M NaCl means amount of moles = 4
lets look at the reaction:
Ag2SO4 + 2NaCl ---> 2AgCl + Na2SO4
notice here that if we were to consume all 4 moles of Ag2SO4, we'd need 8 moles of NaCl for the reaction to proceed
we don't have that much NaCl, we only have 4 moles
so the amount of NaCl will determine how much the reaction happens
in more technical terms we would say that NaCl is the "limiting reagent"
all 4 moles of NaCl will be consumed, and 2 moles of Ag2SO4 will be consumed
the other 2 moles of Ag2SO4 are just extra and put to the side because they cant participate in the reaction
so now
we can see that there will be 4 moles of AgCl formed, and 2 moles of Na2SO4 (since 1 stoichiometric coefficient corresponds to 2 moles in our scheme)
now the total volume given to us is 2L of Ag2SO4 solution and 4L of NaCl solution
meaning that this solution as a whole is 2+4=6L in volume
can go from there?
Ok so in calculating final molarity do we also consider the two moles of Ag2SO4 that were left over?
Or do we just do it for Na2SO4
And since AgCl is a precipitate it won’t have molarity right?
good question
i'd say yes
since it's still in the solution
hasn't gone anywhere, just wasn't part of the reaction
i'm not actually sure for AgCl, getting confused on what "all the ions" means
try and see, and lemme know?
Yeah you don't have to consider AgCl, Precipitation means it won't exist in the form of ions. Consider the excess reagent (4 mol Ag+ and 2 mol SO42-) and the product 2mol So42- and 4 mol Na+.. should yield 2M
Alright thanks dude
+solved @ns @Nimboi
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