Simple Stoic Qustion.........

[bucket]

Its been so long since I've done chem..

"Iron metal is extracted in a blast furnace by a reaction between iron(III) oxide and carbon monoxide: Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)
To produce 1000kg of iron, calculate the mass of the iron(III) oxide required"

I used to find stoichiometry sooo easy  too many beers over the holidays...

[enwiabe]

Alright, so let's look at the molar ratios.

1 Fe2O3: 3 CO : 2 Fe : 3 CO2

You're given that 1000kg of iron is produced (1000kg = 1,000,000 grams)

You know that n = m/M. I assume you're familiar with this equation. If not, n = mole of substance, m = mass (in g) of substance, M = Molar Mass (g/mol) of substance.

 from memory M(Fe) = 55.9 (could be wrong, i don't have a periodic table on me)

therefore, n(Fe) = m(Fe)/M(Fe), therefore, n(Fe) = 1000000/55.9 ~ 17889.088 mol (stored on calculator).

Now, if you have 17889.088 mol of Fe, and for every 1 mol of Fe2O3, you obtain 2 mol of Fe, then to produce 17889.088 mol you need to HALVE 17889.088 mol, to obtain ~ 8944.54 mol of Fe2O3. Because, for every 1 mol of Fe2O3 consumed, 2 mol of Fe is produced. Hence, 8944.54 mol of Fe2O3 produced double the amount (17889.088 mol) of Fe.

Now that you have the mol you've gotta sub back into the equation, n(Fe2O3) = m(Fe2O3)/M(Fe2O3)

we know that n ~ 8944.54, and M = 2 x 55.9 + 3 x 16 = 159.8

therefore, if n = m/M then m = n x M = 8944.54 * 159.8 ~ 1429338g =1429.338kg

However, you're given FOUR sig figs in the question stem (1000 kg) so you give the answer as 1429kg.

[bucket]

i keep getting 714kg as my answer, which is approximately half and i dont see where it would double in my method of working.
This is what I did.

m(fe) in the end is 1000kg.
i go m/M = 1000/112 = 8.93mol

the ratio is 1:2 (Fe₂O₃:2Fe)
so i halve the mol
8.93/2 = 4.46mol

now converting it back to mass
m(Fe₂O₃) = n X m = 4.46 x 160 = 713.6 or 714. around about half the answer.


sorry if its hard to understand my explanation
i havent looked at using latex yet.

[bucket]

oh god thanks lol.
that other guy deleted his post, he was wrong anyway
cheers

[enwiabe]

The book is wrong because it rounded off too early in the question. ALWAYS store values on your calculator. It should be 1429kg not 1430kg.

[bucket]

lol i made such a small and stupid error.
i did 1000kg/112
instead of 1000kg over 55.9.
because it said 2Fe and i was reading it as Fe₂.

geez. i need to go to school.

[Mao]

*kills self with own stupidity*

I earlier posted a method but was defeated in the shame of bearing the incorrect answer and was deleted
i just realised it was a stupid systematic error of pressing "enter" twice on the calculator.... (it ended up adding O₃ twice)

method as follows:

keeping in mind of the molar ratio, work out the mass ratio between the two molecules/compounds








hence to produce 1000kg of Iron, 1430 kg of Iron Oxide is required (i rounded it to 4 sig figs)

ps this answer is different to enwiabe's as I used whereas he used 55.9

[enwiabe]

Ah okay, 55.9 must be rounded. My bad, the book is right!

[Collin Li]

Ah okay, 55.9 must be rounded. My bad, the book is right!

However, VCAA is wrong. They will only give you these values to one decimal place. So do it the "wrong" way

[lanvins]

another stoic question

.5mol of Zn reacts with .5 mol of HCl according to Zn(s) +2HCl = ZnCl2 (aq) H2(g)
Calcualte the amount of hydrogen gas produced, in mol

[Collin Li]

Okay, HCl is the limiting reagent on first glance (because 2 mol of HCl are consumed per mole of Zn, while there is 0.5 mol of Zn and only 0.5 mol of HCl).

This means:

[lanvins]

thanks, so when the moles are the same u use the limiting reactant?

[Collin Li]

You always use the limiting reactant, because that tells you to what extent the reaction proceeded. For example, if you had 0.01 mol of HCl, and 50 mol of Zn, the excess of Zn doesn't matter, you only have 0.01 mol of HCl, so that means only 0.005 mol of hydrogen can get produced.

[bucket]

lol thanks coblin, i was about to ask about this limiting reaction crap.

[kido_1]

Be careful with limiting reagants and determining mole ratios.