Answer:
See Explanations ... I'll have to deliver my explanation in two posts. One is beyond the limits allowed.
Explanation:
Stoichiometry is very easy if you understand the relationship between the 'mole' and the balanced chemical equation. Let's start by defining the 'mole'
The 'mole' of substance is the mass of substance (element or compounds) that contains 1 Avogadro's Number ( = 6.02 x 10²³ ) of particles. The mass of substance that does contain 1 Avogadro's Number of particles is 1 formula weight of substance (also, molecular weight of substance). That is,
1 mole = 1 formula wt. = 1 mole weight = 1 Avogadro's Number = 6.02 x 10²³ particles of substance.
So, assuming you can determine formula weight then that value expressed in terms of grams is a 'Gram Formula Weight'; that is, 1 GFW = 1 mole wt.
At this point it is assumed that you understand how to balance simple chemical equations by inspection. That is, by the Law of Mass Balance ...
total mass of reactants = total mass of products
In problem 1 you are given iron (Fe) + molecular oxygen (O₂) => iron(III) oxide (Fe₂O₃) ... 4Fe + 3O₂ => 2Fe₂O₃ ... ∑mass R = ∑mass P.
Now,<em> focus on the coefficients of each substance,</em> that is, 4 for Fe, 3 for O₂ and 2 for Fe₂O₃. These values (coefficients) are moles of each substance in a special type of equation called the 'Standard Equation'. The Standard Equation is one which when balanced its coefficients are in the smallest whole number mole ratio of substance. That is the equation above is showing 4 moles of Fe, 3 moles of O₂ and 2 moles of Fe₂O₃. This is important in that most balanced equations that will be given in textbook problems will be in Standard Form.
From the Standard Form, the coefficients can be doubled, or divided in half, or whatever multiple or fraction one desires. The reaction would still perform in correct proportions. However, the multiples or fractions would not be in Standard Form. Most, if not all, stoichiometry problems will provide only the Standard Form.
Example, let's take the given equation above (from problem 1)
4Fe + 3O₂ => 2Fe₂O₃.
Both of the following are in correct proportions and will react with respect to the proportions YOU choose ...
(multiply by 2) => 8Fe + 6O₂ => 4Fe₂O₃.
(dividing by 2) => 2Fe + 3/2O₂ => Fe₂O₃.
Both are balanced and conform to the Law of Mass Balance but are not in Standard Form. Oh, and the Standard Form equations are assumed to be at 0°C and 1 atm, even if not specified in the problem. You may not need that now, but will become important in Gas Laws.
Let's move on to stoichiometry calculations. There are several methods being taught within the chemed. community, but I'm going to focus only on using mole ratios.
My approach begins with this rule => <em>convert reaction data to moles, solve using mole ratios and convert to needed dimension at the end of the problem. </em>This will simplify a lot of problems once you understand what it's about.
OK, for problem 1 ...
How many molecules of Fe₂O₃ can be produced from 10.5 moles of O₂?
Given the Standard Form equation 4Fe + 3O₂ => 2Fe₂O₃.
Now, write (neatly) the equation and list the givens under the specific substances ...
4Fe + 3O₂ => 2Fe₂O₃.
ex* 10.5 moles ? molecules note: ex* = excess (not used)
In this problem, since O₂ is already in terms of moles no conversion needed.
What you need to focus on is the relationship between the coefficients of O₂ and Fe₂O₃. Note that the coefficient of Fe₂O₃ is SMALLER than the coefficient of O₂. This means that the number of moles of Fe₂O₃ formed will be smaller than moles of O₂ given, that is 10.5 moles. How much smaller? Easy Calculation => Take the coefficient of O₂ and the coefficient of Fe₂O₃ and make a ratio of the numbers that will make 10.5 SMALLER. That is, you should use 2/3(10.5) to get a smaller number, not 3/2(10.5) because that would give a larger number and inconsistent with the ratio given in the balanced equation.
So, moles of Fe₂O₃ produced = 2/3(10.5) moles Fe₂O₃ = 7 moles Fe₂O₃. Note that 7 moles is smaller than 10.5 moles.
Now, the number of molecules of Fe₂O₃ is also an easy calculation.
Remember that 1 mole contains 6.02 x 10²³ particles, then 7 moles would contain ...
= 7 moles Fe₂O₃ x 6.02 x 10²³ molecules Fe₂O₃/mole Fe₂O₃
= 4.21 x 10²⁴ molecules Fe₂O₃.
Go to 2nd post for problem 2 solution.
