Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars everyday. Although the actual process also requires water, a simplified equation (with rust shown as iron(III) oxide (Fe2O3) is:
4Fe(s)+3O2(g) --> 2Fe2O3(s)
delta(h reaction =-1.65 x 10^3
How much heat is evolved when 0.250 kg of iron rusts?.....KJ MOLES in SCIENTIFIC NOTATION .
How much rust forms when 9.40x 10^3 kJ of heat is released?..... SCIENTIFIC NOTATION ....
answer:
4Fe(s)+3O2(g) --> 2Fe2O3(s)
delta(h reaction = -1.65 x 10^3 kJ
A)How much heat is evolved when 0.250 kg of iron rusts?
Solution :- using the mole ratio of the balanced reaction we can calculate the amount of heat produced as follows.
(0.250 kg Fe*1000 g/ 1kg)*(1mol / 55.845 g)*(-1650 kJ / 4 mol Fe) = -1.85*10^3 kJ
Therefore it will give -1.85*10^3 kJ heat (sign is negative because its exothermic reaction)
So we can write it as amount of heat given = 1.85*10^3 kJ
B)How much rust forms when 9.40x 10^3 kJ of heat is released?
Solution :-
Using the mole ratio of the reaction we can calculate the amount of the rust formed as follows
(9.40*10^3 kJ * 2 mol Fe2O3/ 1.65*10^3 kJ)*(159.687 g / 1 mol Fe2O3) =1.82*10^3 g Fe2O3
Therefore it will give 1.82*10^3 g rust
Am I right? Thank you
Your calculations for both parts A and B are correct.
In part A, you correctly converted the mass of iron (0.250 kg) to moles using the molar mass of iron (55.845 g/mol). Then, using the mole ratio between iron and the heat released from the reaction (-1650 kJ / 4 mol Fe), you calculated the amount of heat evolved as -1.85 * 10^3 kJ. The negative sign indicates that the reaction is exothermic.
In part B, you used the mole ratio between the heat released (9.40 * 10^3 kJ) and the formation of rust (2 mol Fe2O3 / -1.65 * 10^3 kJ). Then, using the molar mass of Fe2O3 (159.687 g/mol), you calculated the amount of rust formed as 1.82 * 10^3 g.
Overall, your answers are correct and in scientific notation. Well done!