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.
<h3>Chemical reaction:</h3>
4 Fe + 3O2 ------ 2Fe2O3
∆H = -1.65×10³kJ
A) Given,
mass of iron = 0.250kg = 250 g
<h3>Calculation of number of moles</h3>
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?
