Stoichiometry

__A + __B --> __AB

1. Multiply the number of moles (A) by the number of moles (# molecules)1 for B.

2. Divide the result by the number of molecules for A.

- Formula

Mol A * # molecules B/# molecules A

- Conversion factor

Mol A * # molecules (mol) B = Mol B

----- # molecules (mol) A

*****1 the number of moles in this case refers to however many molecules of each substance is within the balanced equation (# molecules). For clarity sake I put # molecules in place of "moles" where appropriate; however, on practice problems or demonstrations shown in textbooks, it's likely it will say moles instead of # molecules.*****

__A + __B --> __AB

1. Multiply mol A by # molecules B; divide by # molecules A. (Mol A --> Mol B)

2. Multiply result from 1 by molar mass B (mol B --> Mass B).

- Formula

Mol A x # molecules B/# molecules A

- Conversion Factor

Mol A x # molecules (mol) B x molar mass B = mass B

--------- # molecules (mol) A ----- 1 mol B

__A + __B --> __AB

1. Multiply Mass A by # molecules B.

2. Divide by molar mass multiplied by # molecules A.

- Formula

Mass A x # molecules B/(Molar Mass A x # molecules A)

- Conversion Factor

Mass A x 1 mol A x # molecules (mol) B = mol B

--- molar mass A - # molecules (mol) A

__A + __B --> __AB

1. Convert from grams (g) to mol for substance A (mass A --> mol A).

2. Divide mol A by mol B (mol A --> mol B [Molar Ratio of Substances]).

3. Multiply mol B by molar mass B (mol --> mass)

In summary, you are converting from grams A to mol A, then mol A to mol B, then mol B to grams B.

- Formula

Mass A x # molecules B x Molar Mass B/(Molar Mass A x # molecules A)

- Conversion Factor

Mass (g) A * 1 mol A -- x -- # molecules (mol) B -- x -- molar mass B = mass B

----- molar mass (g) A --x-- # molecules (mol) A --- x --- 1 mol B

The reagent in excess or reactant in excess is the product left over or reagent that is leftover from the reaction; in other words, the reagent that has the most product that did not react.

To identify the limiting reagent or limiting reactant, identify which substance produced the least amount of product (which reactant yields the least amount of product).

Steps (given masses of products):

A. Identify amount of product created per reactant (reactant --> product yield).

1. Balance the equation if it has not been done already.

2. Convert the given masses of reactant (A, B, etc.) to mass product (C) (see "Mass A --> Mass B [Mass --> Mass]" above). (mass reagent --> mass product)

B. Identify the limiting reagent. The limiting reagent (reactant) will be the reactant (A, B, etc.) that yields the least amount of C (product).

C. Identify the reagent in excess. The reagent in excess (reactant) will be the reactant (A, B, etc.) that yields the most amount of C (product).

D. Give how much reagent in excess remain unreacted. (How much reactant is leftover). For simplicity sakes, the limiting reagent will be A and the reagent in excess will be B. "For every X grams of the limiting reagent, there is Y grams of the reagent in excess".

Mass Limiting Reactant --> Mass Reagent in Excess:

First convert the mass of the limiting reagent to the mass of the reagent in excess (mass limiting reagent : mass reagent in excess [reacting] ratio); then subtract the mass of the limiting reagent from the mass of the reagent in excess (that reacted)

1. Convert the mass of the limiting reagent to mass of the reagent in excess (ratio mass limiting reagent: mass reagent in excess). Refer to "Mass A --> Mass B" above.

2. Subtract the original amount of B (reagent in excess) from the amount of B needed to react with A (limiting reagent).

Reagent in Excess leftover = Starting Mass A - Reacting Mass B (step 1 answer)

E. Find % yield.

% yield = actual yield (given)

------- theoretical yield (must be found)*

* the theoretical yield is the amount of product theoretically produced by the limiting reagent; the actual yield is the amount of product actually produced by the reactants; the theoretical yield will have been found in step A. The actual yield will be given within the worded problem.

100.0g Na2CO3 x 1 mol Na2CO3 x 1 mol Fe3O4 x 231.6g Fe3O4

------------------ 106.0g Na2CO3 -- 4 mol Na2CO3 -- 1 mol Fe3O4

= 54.62g Fe3O42

(2 this is known as the theoretical yield, which will be needed when calculating percentage yield later on).

300.0g Fe3Br8 x 1 mol Na2CO3 x 1 molecules Fe3O4 x 231.6g Fe3O4

----------------- 106.0g Fe3Br8 ---- 4 molecules Fe3Br8 -- 1 mol Fe3O4

= 86.12g Fe3O4

B. What is the limiting reagent?

The limiting reagent in this case is Na2CO3 because it has the lowest theoretical yield for producing Fe3O4 (54.62g Fe3O4 vs 86.12g Fe3O4).

C. What is the reagent in excess?

The reagent in excess is Fe3Br8 because it has the highest theoretical yield for producing Fe3O4.

D. How many grams of the reagent in excess(Fe3Br8)remain unreacted?

100.0g Na2CO3 x 1 mol Na2CO3 x 1 molecules Fe3Br8 x 806.8g Fe3Br8

------------------- 106.0g Na2CO3 -- 4 molecules Na2CO3 - 1 mol Fe3Br8

= 190.3g Fe3Br8

300.0g Fe3Br8 - 190.3g Fe3Br8 = 109.7g Fe3Br8 leftover (unreacted)

E. If 42.75g of Fe3O4 were isolated, what is the % yield?

% yield = 42.75g

----------- 54.62g2 x 100% = 78.27%

To solve stoichiometry problems, you must first do two very important things.

1) Write a balanced equation for the reaction.
2) Convert all amounts of products and/or reactants in the question into moles.
To find out how to do both of these see the Related Questions links to the left of this answer. (Note that if the question involves gases, and the amount of gas is given as a volume, you need to use the Ideal Gas Law. How to do that is also listed under the Related Questions).
Once all quantities have been converted to moles and you have a balanced reaction, you are ready to actually use stoichiometry. The idea of stoichiometry is really quite simple. The coefficients (or numbers) in front of each reactant and product in the balanced chemical reaction tells you the ratio of how much of each you will react/produce. Let's take a simple example: the reaction of hydrogen gas (H2) with oxygen gas (O2) to form water (H2O). 2H2 + O2 ---> 2H2O In the balanced reaction, there is a 2 in front of the H2, and although nothing is written, that means that there is really 1 in front of O2, and a 2 in front of H2O. Stoichiometry tells us that because of the way the numbers in balanced reaction came out, we need two molecules of H2 to react with each one molecule of O2, and also, that this will form 2 molecules of water. I can say the same thing using moles: For each mole of O2 I react, I need 2 moles of H2, and I will produce 2 moles of O2. All I'm using is the ratio of the coefficients from the balanced reaction. That is stoichiometry! So, using this I can say things like: -- If I reaction 0.5 moles of O2 completely, I will make 1 mole of water
-- If I made 4 moles of water, then I consumed 4 moles of H2 and 2 moles of O2.
-- If I want to completely react 5 moles of H2, then I need 2.5 moles of O2 (and I will get 5 moles of H2O). Makes sense? The ratio of H2 to O2 to H2O is 2:1:2, and that always holds for this reaction.
Now let's make it a bit more complicated.
--Problem: If I burn 10 grams of methane, how many grams of CO2 will be produced?
-- Answer: As I said, to solve this problem we need two things: a balanced reaction, and to convert all quantities to moles. First, let's write a balanced reaction. In this reaction, methane gas (CH4) gets burned in oxygen (O2) to form carbon dioxide (CO2) and water vapor (H2O). The balanced reaction is:
CH4 + 2O2 --> CO2 + 2H2O
Notice the ratio is now 1:2:1:2 (reading the reaction from left to right). So that tells me for every mole of CH4 I react, I need 2 moles of O2, and I will get out 1 mole of CO2 and 2 moles of H2O.
Now, I need to convert the 10 grams of methane into moles, because stoichiometry only works for moles and NOT grams! So I use the molar mass of CH4, which is 12.011+4*1.0079 = 16.0426 grams per mole. So to convert to moles, I just divide: 10 grams ÷ 16.0426 gram/mole = 0.6233 moles From the stoichiometry, I now know that if I react 0.6233 moles of methane, I will need twice that many moles of oxygen, or 1.2467 moles O2, and I will get 0.6233 moles of CO2 and 1.2467 moles of H2O as products. But the question asked for a number of grams of CO2, not moles. So once I'm done using stoichiometry, I convert back to grams, now using the molar mass of CO2 (which is 12.011 + 2*15.999 = 44.009 g/mol). So 0.6233 moles of O2 is: 0.6233 moles * 44.009 grams/mole = 27.4308 grams of CO2 Notice that the ratio of moles of CH4 to CO2 is 1 to 1, but the ratio of the weights is totally different. Remember, STOICHIOMETRY ONLY WORKS ON MOLES! I can also find how many grams of water I'll produce, just for fun! The molar mass of water is 15.999 + 2*1.0079 = 18.0148 g/mol. So to convert 1.2467 moles H2O to grams: 1.2467 moles * 18.0148 gram/mole = 22.4591 grams of H2O
to answer stoichiometry problems there are several ways.

one way is the mole concept(mol)

-always remember Avogadro's number which is 6.02x10^23.

eg. how many atoms are there in 1 mol of glucose(C6H12O6)?

1 mol C6H13O6 x 6.02x10^23= 6.02x19^23

1 mol C6H12O6

cancel 1 mol C6H12O6 leaving the answer.

When you write a balanced chemical equation, the number of atoms that go into the reaction (as reactant) must equal the number of atoms that come out as products (for each type of atom). To balance a chemical equation, you must use coefficients in front of molecules to make these numbers come out right so that the reaction is balanced. Stoichiometry allows you to use these coefficients to predict how much of a certain molecule you will create for a reaction if you put in a certain amount of reactants. Let me illustrate with an example (which is the burning of propane gas in oxygen to form carbon dioxide and water):

C3H8 + O2 --> CO2 + H2O

This reaction is not balanced though (Look at how many C atoms go in and how many come out. Do the same with O and H. More go in than come out, right?). Instead it should be:

C3H8 + 5O2 --> 3CO2 + 4H2O

This balanced reaction tells us then that for each propane molecule, C3H8, that is burned, it will produce 3 molecules of carbon dioxide, CO2, and also 4 molecules of water, H2O. The ratio is 1 to 3 for carbon dioxide, and 1 to 4 for water. I can also say that if I know that I burned propane, and I produced 8 molecules of water, I know that I must have burned 2 molecules of C3H8.

The ratio is always 1 to 4, just like in balanced reaction above. It is all based on the ratio of the coefficients. I haven't mention oxygen yet, but it's the same thing. If I burn 1 molecule of propane, I'll need 5 of O2 in order for the reaction to work. The ratio here is 1 to 5. I can even say that if I burned propane and I get 5 molecules of water out, than I must have used 5 molecules of oxygen (and 1 molecule of propane, and also I got out 3 molecules of carbon dioxide along with the water!). The ratio is then 1:5:3:4, which is just like in the balanced reaction above. Note that we usually don't talk about single molecules burning, but rather moles of molecules (which is just a whole lot of molecules). The same rules work in exactly the same way with moles. If I burn 5 moles of propane, I know I will produce 15 moles of carbon dioxide and 20 moles of water. Or if I use up 5 moles of oxygen burning propane, I'll have burned 1 mole of propane, and made 3 moles of CO2 and 4 moles of H2O.
it is the study of relative proportions in which substances react or in which elements form compounds.

The relative molecular mass of magnesium chloride is approximately 60. The molar mass is therefore 60g per mole. Therefore there is 0.42mol of formula units in 2.5 grams.

The formula mass of MgCl2 is 24.3 + 2(35.5) = 95.3Amount of MgCl2 = 2.5/95.3 = 0.0262mol

There are 0.0262 moles of formula unit in 2.5 grams of magnesium chloride.

To get the number (not in moles), multiply the amount in moles by the Avogadro's constant.

That depends on which molecule you are referring to when you say "iron oxide". All of the following are correctly referred to as "iron oxide" - either as iron (II) oxide or iron (III) oxide

FeO

Fe3O4

Fe4O5

Fe5O6

Fe5O7

Fe25O32

Fe13O19

Fe2O3

Assuming you are looking to find the number of grams of oxygen and iron respectively required to produce 100 grams of "iron oxide" you would have to refer to the atomic weights of iron (55.845) and oxygen (~ 15.999 or 16) and then use them to find the molecular weight of the chosen form of iron oxide. From that you would calculate the number of grams required from the formula:

For FemOn

g oxygen =

100 g x 1 mole FemOn/([55.845 x m] + [15.999 x n])g FemOn x n moles O/mole FemOn x 15.999 g O/mole O.

g iron =

100 g x 1 mole FemOn/([55.845 x m] + [15.999 x n])g FemOn x m moles Fe/mole FemOn x 55.845 g Fe/mole Fe.