Question

Be sure to answer all parts. Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars every day. Although the actual process also requires water, a simplified equation (with rust shown as Fe2O3) is: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) AH --1.65 x 103 kJ (a) What is the Ahr when 0.250 kg of iron rusts? x 10 kJ (b) How much rust forms when 2.90 x 109 kJ of heat is released? x 10 g Fe2O3

Answer 1

4Fe(s) + 302 (9) - 2Fe2O3(s) 4 Hyxu = -1.65 xlo?ks 4 Hyon is -ve that is the reaction is exollernic reaction, 4 mole ut fe rusts luen Arxn = -1.65x60 kJ YX 3 4856 g af fe rusted then a Hoxu = -1.65x10 ks 0.250kg =250 g of re susted then, 1 A Hyxu = » Т Ahyxu - 250x (-1.65*). - 4x56 - - 1.84 X10 KT ... A Hoxu Fe= during susting of 0.250ks of -1-84 X10 kg b) Fe2O3 - rust Molar mass of rust = ax atauic mass of • Fe' + 3x aшіс мая, с схчем = 2x 56 + 3x16 = 112748 = 160 simol

2 1.65x 10 kJ of heat released - 2 mole of rust forured, de 1.65xlo kJ otheat released 20160 = 320 sot - rust formed, 2.90xl6 KJ of heat released - ? ? Rust Formed = 2.90* 18 x 320 = 562.42ghet 1 1.65x8 562.42 g at rust formed,