Question

Classify the reactions as either endothermic or exothermic.

CALCULATIONS PART A MOLAR MASS OF FelH)(so H0 892.14g-mol RUN 2 Mass of Feau(so. 6H0 eighad out 0-30t0g O.S020 Inital buret reading Cerno) Final burat reading (tnao) Ome Om 8 ml 9.3 m Balancad ryni Mno +sfe 8Ht Hn+5Fe3 +40 Lalaulate concentrotion of the kHnO CHaO) For both runs and calaulate the 0erage Concen tration of Hno. Include unts.

Answer 1

The balanced equation is 2 KMn. Soat 10 Fe (NHG 2 (50) 6H2O + 8 M₂ C0y = 2MM S04 & K₂ SO 4 +68 H20 torea (204)3 + 10 (NH4)2SO4 The molecular weight of Krno y = 158 039 g/mol = 158 g/mol The molecular weight of re (NH4)2 (209) 2.6H20 = 392 g/mol.

:: from the equation we observe that, For (10 x 392) g = 3920 g Mohrs salt (re (NH4)2 (504) 2.6420). the amount of KMnO4 required is (2x 150) g = 316 g. So, for 3920 g Mohr's ealt kon ou required - 316 g i for Ig Mohr's salt Krenda required = 3160 g and for 0.304g Moher's ealt KMnOu required - 16 x 0.304. 3920 3720 g - 96.564 g 3920 g = 0.0245g.

for 0. 3029 Mohes salt KMaou required = 316*0.302 a 3920 g 20.0244 g

o So, in run 1, In 9.3ml komoy the amount of KMnO4 is = 0.0245 g In I ml k nou the amount of knou is = 0.0245 g 9.3 and in 1000 ml KMnou the amount of kom, is = 0.024581000 9.3 to the concentration of Krmon is 2 26349 2 6349/lit 2.634 g / lit 158 g/mol 158 = 0.0167 mol (lit 20.0167 (M)

de , La Qup 2, т. е. км, 4 ны сль, кН 04 2002 44 4 : 1 км, о , #м дь по км, 4 2 0 and in 1000 ml Kinou the amount of KMM OG. 20.0244 81000 g es - 24:41 + - 2 83 8 . На стил на м п ҚР 4 д-2 - 87 4 | El> - 2 - 3 2 | 158 g/mol О. Не 2 mo1 | 20- сте (1)

So, the average concentration of KMnOy is → 0.0167 toote 2 (M) 2 0.0349 (M) 20.01745 (M)