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?
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).
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?
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.
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.