0
__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%
What else can I help you with?
What are the common nature of stoichiometry?
Stoichiometry involves calculating the quantities of reactants and products in chemical reactions, based on the balanced chemical equation. It often deals with mole-to-mole ratios, mass-to-mass relationships, and volume conversions. Stoichiometry is essential for determining the optimal reaction conditions and predicting the outcomes of chemical reactions.
What Fewer steps are required to solve stoichiometry problems when?
Fewer steps are required to solve stoichiometry problems when the given quantities are well-balanced in terms of moles and when the molar ratios in the balanced chemical equation are easy to work with. This simplifies the calculations and reduces the need for additional conversions or adjustments.
How do you solve stoichiometry?
You don't solve stoichiometry. The questions and answers that arise in stoichiometry are merely manipulations of permanent relationships between things (e.g. there are approximately 70.9 grams in one mole of chlorine gas). The conversions needed to report an answer of a stoichiometric problem are the part that take work to overcome mentally. One has to evaluate the units that a value starts with and the units the final answer requires and think about what conversions are needed in between.
What are the major types of stoichiometry problems?
The major types of stoichiometry problems involve calculating the quantities of reactants and products in a chemical reaction. This includes determining mole ratios, mass-mass relationships, limiting reactants, and percent yield. Other common types of problems include volume-volumetric relationships and stoichiometry involving gases.
What are the two kinds of stoichiometry?
The two kinds of stoichiometry are composition stoichiometry, which involves calculating the mass percentage of each element in a compound, and reaction stoichiometry, which involves calculating the amounts of reactants and products involved in a chemical reaction.
What are the common nature of stoichiometry?
Stoichiometry involves calculating the quantities of reactants and products in chemical reactions, based on the balanced chemical equation. It often deals with mole-to-mole ratios, mass-to-mass relationships, and volume conversions. Stoichiometry is essential for determining the optimal reaction conditions and predicting the outcomes of chemical reactions.
What Fewer steps are required to solve stoichiometry problems when?
Fewer steps are required to solve stoichiometry problems when the given quantities are well-balanced in terms of moles and when the molar ratios in the balanced chemical equation are easy to work with. This simplifies the calculations and reduces the need for additional conversions or adjustments.
How do you solve stoichiometry?
You don't solve stoichiometry. The questions and answers that arise in stoichiometry are merely manipulations of permanent relationships between things (e.g. there are approximately 70.9 grams in one mole of chlorine gas). The conversions needed to report an answer of a stoichiometric problem are the part that take work to overcome mentally. One has to evaluate the units that a value starts with and the units the final answer requires and think about what conversions are needed in between.
What are the major types of stoichiometry problems?
The major types of stoichiometry problems involve calculating the quantities of reactants and products in a chemical reaction. This includes determining mole ratios, mass-mass relationships, limiting reactants, and percent yield. Other common types of problems include volume-volumetric relationships and stoichiometry involving gases.
What are the two kinds of stoichiometry?
The two kinds of stoichiometry are composition stoichiometry, which involves calculating the mass percentage of each element in a compound, and reaction stoichiometry, which involves calculating the amounts of reactants and products involved in a chemical reaction.
How do you do stoichiometry and energy problems?
To solve stoichiometry problems, start by balancing the chemical equation. Then, use the mole ratio between the reactants and products to convert between moles of the given substance and the substance you are trying to find. For energy problems, use the appropriate formulas (like Q=mcΔT for heat transfer) and consider the specific heat capacity of the substances involved. Watch for units and conversions when solving both types of problems.
How much C02 is produced from anacroically converting 100 pounds of cellulose to methane?
That's a tough one. There are multiple reaction pathways for methanogenesis using cellulose. If you're given one formula that works though, you have to balance it, then do some molar conversions, then some stoichiometry.
What is an example of stoichiometry?
An example of stoichiometry is determining the amount of product that can be produced in a chemical reaction. For instance, if you have the balanced chemical equation 2H2 + O2 -> 2H2O, and you know you have 4 moles of H2 and 2 moles of O2, you can use stoichiometry to calculate that you can produce 4 moles of H2O.
What is Stoichiometry and non-stoichiometry defect?
Stoichiometry is the relationship between the amounts of reactants and products in a chemical reaction. Non-stoichiometry defects occur when there is a deviation from the ideal ratio of atoms in a compound due to factors like missing or extra atoms, resulting in properties different from those of a stoichiometric compound.
What is the first step in stoichiometry problems is to?
The first step in stoichiometry problems is to write a balanced chemical equation for the reaction you are studying.
Where can one find weight conversions?
A person can find weight conversions in several different places. Some of these places include Online Conversions, Scales Galore, and Metric Conversions.
How can you identify a stoichiometry problem?
Stoichiometry problems involve calculating the quantities of reactants and products in a chemical reaction based on balanced chemical equations. You can identify a stoichiometry problem if you are given information about the amounts of substances involved in a reaction, and you need to determine the amounts of other substances produced or consumed.
