Question

on 8 of 9 > If a solution containing 27.79 g of mercury(II) nitrate is allowed to react completely with a solution containing 7.410 g of sodium sulfate according to the equation below. Hg(NO3)2(aq) + Na, SO, (aq) — 2NaNO3(aq) + Hg50 (9) How many grams of solid precipitate will be formed? mass: How many grams of the reactant in excess will remain after the reaction? mass:

Answer 1

Hg CNO 3/2 (aq) + Na soulag) 2 NaNo, (aq) + Higson (s) Giren mass of o mass of Giren malor meiss of mass of Hg (NO 3 ) = 27.799 Hg (NO) = Afanie mass of fg+ 2 x Alanie mass of N +6X Atence mass of a = 200+ 9x14 + 6x16 = 200+ 28+96 = 324 glmal Na sou At 9.410g Na, so y = 2x Atemie mass of Nat Atomie mass of s +4x Atamie mass of o = 2x 23+ 32 + 4x16 - 46 +32 + 64 = 142 gimal males of Halog) = Giren mass = 27.99 = 0.086 Molar mass 324 males of Males Nay sou = 7.410 7.410 = 0.0522 o. 0522 142 mass So Nag soy has les no. of maler at given So Na, son is limiting reagent. sold precipitate ferm careepond to Nasoy. I male Masom produces = imale Higsoy es ) 0.0522 male of Nas son produces = 0.0522 male Hason Masou cs sc 901 male 00522 male Scanned with CamScanner mass precipitale Mason - malar maw of Mason Precipitate nggon 296 glmal

00522 x 296 = nasoy precipitate 15.45g & Higson poreupitate So, solid freupitate formed = prysg 1 Hg(NO3) cag) remain in the reaction as it was in eney. 0.0522 completly react with 0:0522 0.0522 0:0522 male of HgNo male of Magson but we have 0086 male of Hq Alog) Remaining Hg (Nwg) = = 8.686- 0.0522 0.0338 male mass of Mg(NO3) = 80336 x Malar mass of Hg Nog), = 0.0378 x 334 g made = 10.95 g 10.95 g / reactent ( ngo ) remain difter the reaction cs dome Scanneauvith CamScanner