Question

Fill in the Blanks A solution containing 58.0 g of mercury(II) nitrate is allowed to react completely with a solution containing 16.642 g of sodium sulfate according to the equation below: Hg(NO3)2(aq) + Na2SO4(aq) + 2NaNO3(aq) + HgSO4(s) How many grams of solid precipitate will be formed if the reaction has a 100% yield? Solid precipitate grams How many grams of the excess reagent will remain ilter the Excess reagent remaining grams If the reaction has only an 80% yield, how many grams of solid precipita Solid precipitate

Answer 1

324 Hg(NO3) caq) + Naz Soplaq) Hg Sou (s) + 2NaNo caq Imol inst imole linol 296 g g 324.6 gole 142 Given mass of Hg(NO3)2 = 58.00 mass of ways = 16.642g Let as determine Limiting reagent fromaan on 2 49 Hej (NO3)2 gives 296.68g Hg say 589 » » ? = 58 x 296.65_53.1g Hgscy ' 142g nasa gives 296.65 g Hgrey - = 16-642 x 296.65 142 ... = 34.779 su forming 34 ng Hgs 4 Na sy is limiling reagent and solid precipitale Haso = 34.79 Excess reagent : 1429 Na s need 324.6g Hg (823) 16-6429 = 16.642 4324.6 - 38.04g Hg (N93) 142 Excess reagent #ging he remaind -58-38.04 = 19.069 so percent yield - Actual yild x 100 Theoretical field. Actualyidd - 80 x 34.71, 27.8169 :: 16.6429 .: As Nasou forming 34-ng to