Am I correct?
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
A) Wow, 0.250 kg of iron rusts and gives off -1.85*10^3 kJ of heat? That's quite the hot mess!
B) And if 9.40*10^3 kJ of heat is released, it forms 1.82*10^3 g of rust. Well, I guess that's one way to leave your mark on the world, with a lot of rusty iron!
Yes, you are correct. The amount of heat evolved when 0.250 kg of iron rusts is -1.85 * 10^3 kJ, and the amount of rust formed when 9.40 * 10^3 kJ of heat is released is 1.82 * 10^3 g.
To calculate the amount of heat evolved when 0.250 kg of iron rusts, you can use the mole ratio from the balanced equation and convert the mass of iron to moles:
1. Convert the mass of iron from kilograms to grams: 0.250 kg * 1000 g/1 kg = 250 g.
2. Use the molar mass of iron (Fe) to convert grams of iron to moles: 250 g Fe * (1 mol Fe / 55.845 g Fe) = 4.47 mol Fe.
3. Use the mole ratio from the balanced equation to relate moles of iron to moles of heat released: 4 mol Fe * (-1650 kJ / 4 mol Fe) = -1650 kJ.
4. The negative sign indicates that the reaction is exothermic, so the amount of heat evolved is 1650 kJ.
5. Expressing it in scientific notation, the heat evolved when 0.250 kg of iron rusts is -1.65 x 10^3 kJ.
To calculate the amount of rust formed when 9.40 x 10^3 kJ of heat is released, you can use the mole ratio from the balanced equation to calculate the moles of iron(III) oxide:
1. Use the given amount of heat released: 9.40 x 10^3 kJ.
2. Use the mole ratio from the balanced equation to relate moles of heat released to moles of iron(III) oxide: 9.40 x 10^3 kJ * (2 mol Fe2O3 / 1.65 x 10^3 kJ) = 11.35 mol Fe2O3.
3. Use the molar mass of iron(III) oxide (Fe2O3) to convert moles of Fe2O3 to grams: 11.35 mol Fe2O3 * (159.687 g Fe2O3 / 1 mol Fe2O3) = 1811.55 g Fe2O3.
4. Expressing it in scientific notation, the amount of rust formed when 9.40 x 10^3 kJ of heat is released is 1.82 x 10^3 g.