Question

→ X COD Data: & p 029 A 25.0 mL volume of 0.235 M Sr(NO3)2 is added to 20.0 mL 0.300 M (NH4)3PO4. What is the maximum amount of solid product that can be collected? XI FIRE, ALT+F10 (PC) ER ALT+FN+F10 (Mac). TT TT Arial 3 (12pt) E.E. T. % DO Q E SE T' T, mq - - - O fx Mashups - 16 @ @ % :p 105 10 A 25.0 ml volume of 0.235 M Sr(NO3)2 is added to 20.0 mL 0.300 M (NH4)3PO4. One of the ions in the formula of the precipitate remains in excess. What is the concentration of this ion in mol/L? TTTT Arial 3 (12pt) - E

Answer 1

351(NO3)2 + 2 (NH4)3PO4 = S, (poc), + 6 NH, NO3 20 me 0.300 M (N Hugpon: 6X10% mele 25 me 0.235 M [S(NO3)2] = 5.875x1ömule . Here, 3 ne, 3 mole mole Sn (NO3)2 snl neact with 2 male (NH4)₂004. : . 5,875x103 . . ? x 5.875x173 = 3.92 x 103 male (NH,), Pºt - The limiting neagentSn(NO3)2 The excess reagent = (NH4)3 Pou beartaid The amount of (NHa); po = 3.928103 mole : The amount of (NH4)3 pon excess = (6X163 - 3.92x153) – 2.083x103 = (20+25)=45 me solution contain 2:083x103 mole i mole 1000 11 1 1 = 2:083x103. x 1000 = 0.0463 (M) ... The concentration of exeen - Xeen =0.0463 • mole/L" Ans " 3 mole Sn (NO3) romanetos vint Produce 1 mole Srz (pod. 5.875x173 5.875x10° mele .= 1.958X163 mole प - The molecular weight = Srz (po2 = 452.89) mole 2. The amount of solid = 452.8x1.958x13 0.8878] AS