Posted by Sarah on Friday, November 21, 2008 at 7:37pm.
Watch carefully because this can be confusing. Look carefully at what the question asks.
How many moles and grams of Na2SO4 are in the reaction?
1.55 x 10^-3 mol Na2SO4 is correct. I obtained 0.2206 g which I would round to 0.221 for grams Na2SO4. Note the questions asks for g Na2SO4 in the REACTION.
How many moles and grams of Pb(NO3)2 reacted in the mixture?
You are correct with 1.55 x 10^-3 mols Pb(NO3)2 and 0.514 g Pb(NO3)2 but note that the question asks for mols and grams that REACTED. It didn't ask for mols and grams Pb(NO3)2 that are present in the original 1.009 grams sample. Since Na2SO4 is the limiting reagent, that limits the amount of Pb(NO3)2 that reacts BUT some additional Pb(NO3)2 CAN be present.
What is the percent by mass of each salt in the mixture?
%Na2SO4 = (mass Na2SO4/mass sample)*100 = (0.2206/1.009)*100 = approximately 20%.
%Pb(NO3)2 = (mass Pb(NO3)2/mass sample)*100 = [(1.009-0.221)/1.009] = about 80%
Note that you could have determined % Pb(NO3)2 simply by 100 - %Na2SO4.
But also note that you CANNOT determine %Pb(NO3)2 by (0.514/1.009)*100 BECAUSE the 0.514 g is the amount of Pb(NO3)2 that reacted and not the amount of Pb(NO3)2 in the sample at the beginning. If you do it this way you get about 51% and obviously the two percentages must add to 100%.
•0.471g / 303.26g/mol = 1.553x10^-3 mol PbSO4
• moles of Na2SO4 = 1.553x10^-3 —> 0.2206 g (total in the sample)
• moles of Pb(NO3)2 reacting = 1.553x10^-3 —> 0.51435g
• 1.009 g - 0.2206 g = 0.7884 g Pb(NO3)2 total in the sample
• % Na2SO4 in the sample was: (100)(0.2206 g)/1.009g = ___?
• % Pb(NO3)2 in the sample = [setup] = ___?
That makes sense! I was confused because my answers werent adding up to 100% but you explained it. Thank you!!
What is the molecular form of the equation for the reaction? and what is the net ionic equation for the reaction? Thanks!!
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