posted by Ann on .
a 1.009g mixture of the solid salts Na2SO4 and forms an aqueous solution with the precipitation of PbSO4. The precipitate was filtered and dried and its mass was determined to be .471 g. the limiting reactant was determined to be Na2SO4. Write the molecular form of the equation for the reaction?
Write the net ionic equation
how many moles and grams of na2so4 are in the reaction mixture.
how many moles and grams of pb(no3)2 reacted in the reaction mixture?
what is the percent by mass of each salt in the mixture?
The molecular equation is
Pb(NO3)2(aq) + Na2SO4(aq) ==> PbSO4(s) + 2NaNO3(aq)
net ionic equation is
Pb^+2(aq) + SO4^-2(aq) ==> PbSO4(s)
You know you have 0.471 g PbSO4. Convert that to moles PbSO4, then convert to moles Pb(NO3)2 (since that was the limiting reagent). Convert moles Pb(NO3)2 to grams Pb(NO3)2 which you must have had initially. Then
mass percent Pb(NO3)3 = (grams Pb(NO3)2/mass sample)*100 = ??
1.009 g sample - g Pb(NO3)2 = grams Na2SO4 and convert that to mass percent the same as above.