Question

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: 4Fe(s) + 3O2(g) --> 2Fe2O3(s) Heat of reaction = -1.65x10^3 kJ

(a)What is the delta Hrxn when 0.250 kg of iron rusts?

__x10^__kJ

(b)How much rust forms when 8.000x10^3 kJ of heat is released?

__x10^__gFe2O3

Answer 1

4Fe(s) + 3O2(g) --> 2Fe2O3(s) Heat of reaction = -1.65x10^3 kJ

m = 250 g o FFe

MW Fe = 55 g/mol

mol = 250/55 = 4.545 mol of Fe

4 mol ---> -1.65*10^3

4.545 --> 4.545/4* -1.65*10^3 = -1874.81 kJ

b)

if Q = 9*10^3 kJ

then

4Fe(s) + 3O2(g) --> 2Fe2O3(s) Heat of reaction = -1.65x10^3 kJ

2 mol of Fe2O3 --> -1.65x10^3 kJ

x mol --- > 8*10^3 kJ

ratio (8*10^3)/(1.65*10^3)*2

9.69696 mol of Fe2O3

MW of Fe2O3 = 159.69

mass = mol*MW = 159.69*9.69696 = 1548.5075 grams of Fe2O3