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Help with moles/conc type question

1.00g of lead(II) nitrate is mixed with 129mL of 0.103moldm-3 of sodium sulfate solution.

a) Write a balanced equation = Pb(NO3)2 + Na2SO4 ----> 2NaNO3 + PbSO4
b) what is the limiting reagent? (Do i just calculate the moles for each reactant and the one with lower moles is the limiting reactant?)
c) what is the concentration of sulfate ion that remain in the solution after the reaction is complete, convert your answer from moldm-3 to ppm
d) what is the concentration of nitrate ion that remain in solution after the reaction is complete?

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Which part do you need help with and what have you tried so far??
Original post by Radioactivedecay
Which part do you need help with and what have you tried so far??


I guess i've done b where I got lead nitrate as my limiting reagent so QC and QD, for C I tried to subtract the moles of sodium sulfate and lead nitrate and then just divide the volume (129mL) but don't know if that is right
Thats correct for part B, the element with the fewer moles would be the limiting the agent. For C, try calcaulating the moles of sulfate ion formed using the ratios in the equation and simply divide by the volume(careful with units).
Original post by Radioactivedecay
Thats correct for part B, the element with the fewer moles would be the limiting the agent. For C, try calcaulating the moles of sulfate ion formed using the ratios in the equation and simply divide by the volume(careful with units).


This is my working for C:
Moles of Pb(NO3)2 = 3.02x10^-3
Moles of Na2SO4 = 129/1000 x 103 = 0.013287
Moles of SO4 2- left over = 0.013287 - 3x10^-3 = 0.010268
moles/vol = conc therefore 0.010268/0.129 = 0.079594moldm-3
however they want answer in ppm so do I just multiply by 1000 and and the molar mass of Na2SO4?
Simply convert your answer to milligrams per liter, so ya multiply by the mr and then divide by 1000 not multiply.
Original post by Radioactivedecay
Simply convert your answer to milligrams per liter, so ya multiply by the mr and then divide by 1000 not multiply.


okay cool that gives me 7.5ppm, btw how would i do D?
No nitrate ions reacted in the equation so work out the moles of NaNO3 and that would be the moles of NO3 ions then simply divide by the volume.
Original post by Radioactivedecay
No nitrate ions reacted in the equation so work out the moles of NaNO3 and that would be the moles of NO3 ions then simply divide by the volume.


which volume? the 129mL?
Yup.
Original post by Radioactivedecay
No nitrate ions reacted in the equation so work out the moles of NaNO3 and that would be the moles of NO3 ions then simply divide by the volume.


There was one more question asking to find the concentration of Sodium ions left after the reaction is complete how would i go about doing this?
Is this part of the same question?
Original post by Radioactivedecay
Is this part of the same question?


yeah but i forgot to include it mb
In that case work out the moles of Na2SO4 left after the reaction and double it as it is Na2 to find the moles of Na ions and simple divide by the volume.
Original post by Radioactivedecay
In that case work out the moles of Na2SO4 left after the reaction and double it as it is Na2 to find the moles of Na ions and simple divide by the volume.


the 129mL volume?
yup
Moles deserve hell they will be the reason i fail my chem gcse's. I hope avagadro is in hell.
is this gcse ? or alevels
Original post by IPFromTheEast
is this gcse ? or alevels


undergrad
what is undergrad ?? higher than gcse or ??

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