Question

If a solution containing 19.71 g of mercury(II) chlorate is allowed to react completely with a solution containing 6.256 g of sodium dichromate, how many grams of solid precipitate will be formed? mass: How many grams of the reactant in excess will remain after the reaction? Assuming complete precipitation, how many moles of cach ion remain in solution? If an ion is no longer in solution, entera xero (O) for the number of moles. CIO, C1,07 hp

Answer 1

Balanceel Reaction : - Hg (CI) + Nacron (as) Hg C&O + 2 Nacion (as) polo mass mw chiorate = 19.719 - = 367.45 ym @ mercuric 0.0536 moles. 6 sodium dichromate = 6.256 g = 261.97 Dm 0:0238 mories & Limiting Reagent : In balanced seaction. I mole mescurry chcrate is reacted with Imole Nag cao. Cratio 1:1). Thesefest. o.o536 mole on oso must reacts with 6.0536 mole Na2C%₂0s. But, these py pess amount (6.0238 mop) of 4 (% Oy present & will be completely consumed during reaction thesefex 'No, cron is limiting Reagent In basanced Reaction, amole Haz croy Climitin produce more gerady (Precipitate), cratio 1:0 Therefore, 0.0238 mok Nazar og muss produce 0.0238 mol Hg (87.

Thesefoot mass of precipitates = moles y molecular mass. = 0.0238 mole x 416 588im = 9.948 paims 1 mass & precipitater = 9.948 g) & amount of Hg (C103) unreacted - Total mokes – mores, reacted = 0.0536 - 0.0238 mon = 0.0298 mole i.e. 0.0298 mole Hoo remains in soin a dissociateed as - Hg (C103) T 4g 2+ + 2 c103 mole Hg Caoo) will produce Imole Hg IT (12) & a mole clog (1:2), Thesefox. 2 0.0298 mole Hg Caosz will produces 0.0298 mole Harta 2x 0.0298 = 0.0596 molo 103. amount B Higt in inson = 0.0298 mole - 0.0596 mola

6 Now, Imole Hagcro produces 2 mole noclo, (ratio 1:2) 0.0238 mole Na₂Cron produces 2 x 0.0238= 0.0476 mole Noclo3. dissociated as Now, Imole No CIO Nacio T Nat & 10 I mole I mole I mole : 0.0476 mole noclos will produce 0.0476 mola Nat & 0.0476 mole cloi Hg2+ = 0.0298 mole = 0.0476 poole Not = 0.0476 mole (2077- = 0.0546 mole mass t preci pitates = 9.948 grams)