Question

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ
(b) How much rust forms when 4.85X10³ kJ of heat is released?

Answer 1

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

Chemical reaction:

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

Calculation of number of moles

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

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DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?