Unbalance question ,Anyone help me out ,please ?
The following unbalanced reaction is called the thermite reaction. It releases tremendous amounts of energy and is sometimes used to generate heat for welding.
Al + Fe3O4 ---> Fe + Al2O3
Determine the masses of all the substances present after the reaction if 164 g of Al and 847 g of Fe3O4 react to completion. Some amounts may be zero (0).
Mass of Fe produced? .............g
Mass of Al2O3 produced?...........g
Mass of Al remaining?...........g
Mass of Fe3O4 remaining?..........g
- MaviGozlerLv 59 years agoFavorite Answer
So the first thing you need to do is balance mass and charge in ANY chemical reaction. (Charge will not be really a factor here.
Go from left to right in looking at atoms and their counts. Here's the logic:
1) 1 atom Al on left-hand side (LHS), 2 atoms Al on RHS. Put 2 in front of Al on LHS to balance. Result: 2 Al + Fe3O4 => Fe + Al2O3
2) 3 atoms Fe on LHS, 1 on RHS. Put 3 in front of Fe on RHS.
Result: 2 Al + Fe3O4 => 3 Fe + Al2O3
3) 4 atom O on LHS, 3 on RHS. Put 3 in front of Fe3O4, and 4 in front of Al2O3.
Result: 2 Al + 3 Fe3O4 => 3 Fe + 4 Al2O3
4) Oxygen atoms balance now, but we lost the balance on Fe and Al atoms, so see if you can balance those without changing the oxygen balance. We have 2 Al on LHS, and 4*2=8 Al on RHS. So put 8 on LHS.
Result: 8 Al + 3 Fe3O4 => 3 Fe + 4 Al2O3
5) 9 Fe atoms on LHS, and 3 Fe on RHS. Make Fe on RHS 9
Result: 8 Al + 3 Fe3O4 => 9 Fe + 4 Al2O3 Hurray! every atom and its numbers balance now!
Whew! Now usually the balancing takes fewer steps, but the molecule Fe3O4 is unusual since Fe is not an integer charge on Fe. Since O has 2- and there are 4 for total 8-, and there are 3 Fe, then Fe has a 8/3+ charge. There is no such thing, so since the value is between 2+ and 3+, that means Fe3O4 as a species is a mixture of Fe[2+] and Fe[3+] atoms.
To complete the work, you need FW or atomic weight of all species:
Al: 26.98154 g/mol
Fe3O4: 55.847 g/mol
Fe: 55.847 g/mol
Al2O3: 101.96 g/mol
How many moles of Al are there? 164 g Al * (1 mol Al / 26.98154 g Al) = 6.09245524 mol Al
Moles of Fe3O4: 847 g Fe3O4 * (1 mol Fe3O4 / 55.847 g Fe3O4) = 15.1664369 mol Fe3O4
Determine which reactant is limiting:
1) 6.09245524 mol Al * (3 mol Fe3O4 / 8 mol Al) = 2.28467071 mol Fe3O4
2) 15.1664369 mol Fe3O4 * (8 mol Al / 3 mol Fe3O4) = 40.4438317 mol Al
Eqn 1 shows you need 2.28 mol Fe3O4 to react with the Al you have, and you have 15+ moles. This normally means Al is limiting
Eqn 2 tests the other reactant: if you have 40.4 mol Al, you can react all the Fe3O4, but you have only 6+ moles of Al.
So we confirm that Al is limiting, and thus ALL of it will be consumed and you will have Fe3O4 remaining.
Now calculate the products based on all the Al reacting:
1) Fe: 6.09245524 mol Al * (9 mol Fe / 8 mol Al) = 6.85401214 mol Fe * 55.847 g/mol = 382.776016 g Fe
2) Al2O3: 6.09245524 mol Al * (4 mol Al2O3 / 8 mol Al) = 3.04622762 mol Al2O3 * 101.96 g/mol = 310.593368 g Al2O3
The Fe3O4 remaining?
15.1664369 mol Fe3O4 - ( 6.09245524 mol Al ) * (3 mol Fe3O4 / 8 mol Al ) = 12.8817662 mol Fe3O4 * 55.847 g/mol = 719.407996 g Fe3O4
Put it together now:
Mass of Fe produced: 383 g
Mass of Al2O3 produced: 311 g
Mass of Al remaining: 0 g (remember it was limiting and thus all consumed)
Mass of Fe3O4 remaining: 719 g
Note that I in intermediate calculations, I never round off to significant digits. Do that with the final result. Never round off with intermediate calculations.
Most people are befuddled as to how to set up chemical calculations properly. The way to do it is "dimensional analysis." Every calculation in chemistry can be set up as a multi-factor product in which the units or dimensions of the denominator and numerator cancel and the final units or dimensions should be the target value wanted, such as "grams Fe" or "moles Al2O3". Once the product is set up properly and dimensions cancel, only the arithmetic on the numbers needs to be done.
Also if there is a chemical equation: z A + y B ==> w C + v D, it can be said inter alia that v moles of D are formed from z moles of A.
So if given 54 g of A, and it is converted to its mole amount (say 6.5 moles), then assuming all A is consumed, the number of moles of D formed = 6.5 mol A * (v mol D / z mol A). Notice how the "mole A" dimension cancels out of the product, and only "mol D" remains, its value (6.5*v/z).