Conversions of stoichiometry

Answer 1

Stoichiometry Conversions Using Balanced EquationsMol A --> Mol B (Mol -->Mol)("How many moles of B are needed to react with X mol A?")

__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.*****

Mol A --> Mass B (Mol --> Mass)("How many grams B are needed to produce X mol A?")

__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

Mass A --> Mol B (Mass --> Mol)("How many mol of B are needed to react with X g A?")

__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

Mass A --> Mass B (Mass --> Mol --> Mass)("How many grams of A can be produced from X grams B?")

__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

Limiting Reagent and Reagent in ExcessThe limiting reagent or limiting reactant is the substance that limits the reaction. ("What substance has the least amount produced from a reaction?").

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.

Ex. 4Na2CO3 + Fe3Br8 --> 8NaBr + 4CO2 + Fe3O4 A. How many grams of Fe3O4 can be produced from 100.0g Na2CO3 and 300.0g Fe3Br8?

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%