It connects the reactants and the products in a balanced chemical equation to somehow find a certain aspect of the reactants or the products for example: the mass of a product.
REACTANTS --> PRODUCTS
The base of all stoichiometry is the MOL BOX!
What is the mol box, you ask?
This is a mol box:
REACTANTS --> PRODUCTS
The base of all stoichiometry is the MOL BOX!
What is the mol box, you ask?
This is a mol box:
MM?
This is a shortening of the term Molar mass, molar mass is the mass of a substance or elemnt per mol of this substance.
To find the Molar mass, you have to read the periodic table and add the molar masses of the different elements.
This is a shortening of the term Molar mass, molar mass is the mass of a substance or elemnt per mol of this substance.
To find the Molar mass, you have to read the periodic table and add the molar masses of the different elements.
Here is an example
![Picture](/uploads/1/3/9/9/13992289/1294486.gif?98)
The molar mass of C (carbon) is 12.01 g/ 1 mol of Carbon.
If your balanced chemical equation has C₂, the molar mass of the C₂ would be:
24.02 g/ mol of Carbon
*This is because there is 12.01 grams per 1 mol of Carbon so there must be 24.02 grams per 2 mols of Carbon.
If your balanced chemical equation has C₂, the molar mass of the C₂ would be:
24.02 g/ mol of Carbon
*This is because there is 12.01 grams per 1 mol of Carbon so there must be 24.02 grams per 2 mols of Carbon.
MR?
This is a shortening of the term Molar Ratio, the molar ratio is determined by using the coeffecients before each of the reactants and the products in a balanced chemical equation.
To find the Molar Ratio we must look at the balanced chemical equation.
This is a shortening of the term Molar Ratio, the molar ratio is determined by using the coeffecients before each of the reactants and the products in a balanced chemical equation.
To find the Molar Ratio we must look at the balanced chemical equation.
HERE IS AN EXAMPLE
Consider the balance chemical equation:
N₂ (g) + 3H₂ (g) --> 2NH₃ (g)
This can be expressed as, 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2) to form 2 moles of ammonium (NH3).
The Molar Ratio of Nitrogen to Hydrogen is 1/3
The Molar Ratio of Nitrogen to Ammonium is 1/2
The Molar Ratio of Hydrogen to Ammonium is 3/2
Examples:
2H20 --> 2H2 + O2
1 Step problems: MOLA --> MOLB
15.5 mol O2 --> ? mol H20
15.5 mol 02 X 2 mol H20/ 1 mol O2 = 31 mol H2O
2 Step Problems MolA --> gB
10 g H2 --> ? mol O2
** mm H2 = 2g/mol
10g H2 X 1 mol H2/ 2 g H2 X 1 mol O2/ 2 mol H2O = 2.5 mol 02
3 Step Problems MolA --> MolB
This is the same as the 2 step problem but instead of using only 1 molar mass you must use 2 molar masses of the original substance and the final one.
The only way to understand stoichiometry is to do a lot of practice questions.
N₂ (g) + 3H₂ (g) --> 2NH₃ (g)
This can be expressed as, 1 mole of nitrogen (N2) reacts with 3 moles of hydrogen (H2) to form 2 moles of ammonium (NH3).
The Molar Ratio of Nitrogen to Hydrogen is 1/3
The Molar Ratio of Nitrogen to Ammonium is 1/2
The Molar Ratio of Hydrogen to Ammonium is 3/2
Examples:
2H20 --> 2H2 + O2
1 Step problems: MOLA --> MOLB
15.5 mol O2 --> ? mol H20
15.5 mol 02 X 2 mol H20/ 1 mol O2 = 31 mol H2O
2 Step Problems MolA --> gB
10 g H2 --> ? mol O2
** mm H2 = 2g/mol
10g H2 X 1 mol H2/ 2 g H2 X 1 mol O2/ 2 mol H2O = 2.5 mol 02
3 Step Problems MolA --> MolB
This is the same as the 2 step problem but instead of using only 1 molar mass you must use 2 molar masses of the original substance and the final one.
The only way to understand stoichiometry is to do a lot of practice questions.