Basic introduction to Strychometry. I will teach you all the important Strychometry conversions using my tricks. Firstly, let me teach you that what is Strychometry. Well, the word Strych means element and the word Metry means measurement. So Strychometry is the calculation of products and reactants and a chemical reaction.
For example, consider this general chemical reaction. With the help of stoichiometry, we can find the number of moles of reactants and the number of moles of products. So remember that stoichiometry is the calculation of products and reactants and a chemical reaction.
Now let me teach you the important concept of coefficients and a chemical reaction which a lot of students are not understanding. For example, consider this chemical reaction. Hydrogen gas react with nitrogen gas to form ammonia NH3. Now I will balance this chemical reaction. There are 2 hydrogen atoms and the reactants.
Well there are 3 hydrogen atoms and the products. I place 2 with products and 3 with the reactants. Now 3 into 2 is equal to 6 atoms.
2 into 3 is equal to 6 atoms. There are 2 nitrogen atoms and there are also 2 nitrogen atoms. two nitrogen atoms.
So this is complete balanced chemical reaction. Now this three with hydrogen gas is known as coefficient of hydrogen gas. Secondly, this one with nitrogen gas is known as coefficient of nitrogen gas.
While this two with NH3 is known as coefficient of NH3. Here I will teach you three different stories of coefficients. Firstly, The coefficients represent the amount of substances. For example, this 3 coefficient of hydrogen gas shows 3 moles of hydrogen gas. 1 coefficient of nitrogen gas represents 1 mole of nitrogen gas.
And 2 coefficient of NH3 represents 2 moles of NH3. Secondly, coefficients represent ratio of reactants to products. For example, I write hydrogen gas ratio to nitrogen gas to NH3.
We know that the coefficient of hydrogen gas is 3, that of nitrogen gas is 1 and that of NH3 is 2. Hence this is the ratio of reactants to products. Thirdly, coefficients represents number of molecules. Let I write 3 hydrogen gas, 1 nitrogen gas to NH3.
Now this 3 coefficient of hydrogen gas represents 3 molecules of hydrogen gas. This 1 coefficient of nitrogen gas represents 1 molecule of nitrogen gas. And this 2 coefficient of NH3 represents 2 molecules of NH3.
Thus remember that story 1 tells that the coefficient represents the number of moles. Story 2 tells that the coefficient represents the number of moles. that the coefficient represents the ratio of reactants to the products. Story 3 tells that the coefficient represent the number of molecules. Now, how can we use these stories to calculate the amount of reactants and products?
Well, consider this question. How can you form 6 molecules of NH3? The answer is simple.
According to the second story, the ratio of hydrogen to nitrogen to NH3 is 3 to 1, 1 to 2. We need 6 molecules of NH3. I just multiply 2 into 3. I also multiply the ratio of N2 by 3 and that of H2 by 3. Now 2 into 3 is equal to 6. 1 into 3 is equal to 3. 3 into 3 is equal to 9. This reveals that we need 9 molecules of hydrogen gas. to react with three molecules of nitrogen gas in order to form six molecules of NH3. Let me repeat this important statement. We need nine molecules of hydrogen gas to react with three molecules of nitrogen gas in order to form six molecules of NH3.
To conclude this whole concept, any coefficient in a reaction shows either number of moles, a ratio, are number of molecules and snorted down all these important concepts. Now let me teach you the different conversions of stoichiometry like mole to mole conversion. Consider this problem. How many moles of nitrogen gas is needed to react with 13.5 moles of hydrogen to form NH3? Well, I write the balanced chemical reaction.
We know that hydrogen gas plus nitrogen gas react together to form NH3. I put here 3, 1 and 2. This is the complete balanced chemical reaction. Now according to the given statement, 13.5 moles of hydrogen gas is given.
We need to find the number of moles of nitrogen gas. Although, there are several ways to calculate it. But I will teach you my way to calculate such type of questions.
Firstly, I write ratio of hydrogen to the ratio of nitrogen gas. We know that it is 3 to 1. Now listen carefully. We are already given the number of moles of hydrogen.
I will write this 13.5 moles below the hydrogen gas. Let me repeat it. We are already given the number of moles of hydrogen gas.
I will write. this 13.5 moles below this hydrogen gas. Here if 3 moles of hydrogen gas react with 1 mole, then 13.5 moles of hydrogen gas react with x moles. Now I will just cross multiply them. 3 into x is equal to 1 into 13.5.
I divide both sides by 3. After calculation, I get x is equal to 4.5 moles. Thus, we need 4.5 moles of nitrogen to react with 13.5 moles of hydrogen. Hence, note it down this important conversion.
Secondly, consider this problem. How many moles of sulfur trioxide will form when 8.5 moles of sulfur dioxide react with oxygen? Firstly, I write the balanced chemical reaction.
According to the given statement, Sulfur dioxide plus oxygen gas react together to form sulfur trioxide. Now two oxygen atoms plus two oxygen atoms is equal to four oxygen atoms. While in the products, there are three oxygen atoms.
I place here two. In the product, there are two sulfur atoms. In the reactant, I place here two. Hence this is the complete balanced chemical equation.
According to the given statement, 8.4 moles of sulfur dioxide will react with oxygen to form x moles of sulfur trioxide. Now, I will use my personal way to calculate the number of moles of sulfur trioxide. I established ratio between sulfur dioxide and sulfur trioxide.
We can see that it is 2 ratio to 2. The number of moles of sulfur dioxide is 8.4 moles. We need to calculate the number of moles of sulfur trioxide, let it is x. Now I cross multiply them. 2n2x is equal to 2n2 at 0.4.
I divide both sides by 2. After calculation, I get 8.4 moles. Thus 8.4 moles of sulfur dioxide will react with excess of oxygen to form 8.4 moles of sulfur trioxide. Hence, note it down this important problem. The second type of stoichiometric conversion is mole to gram conversion.
For example, consider this problem. Firstly, I write the complete balanced chemical reaction. According to the given statement, propane C3H8 reacts with oxygen to form carbon dioxide plus water.
It is a simple combustion reaction. Now there are 3 carbons in the reactants and 1 carbon in the products. I place here 3. There are 8 hydrogen in the reactants and 2 hydrogen in the products.
I place here 4. 2 x 4 is equal to 8 hydrogen atoms. Now 2 x 3 is equal to 6 oxygen atoms. Plus 4 oxygen atoms is equal to 10 oxygen atoms.
I write here 5. Hence, this is the complete balanced chemical equation. Now, 4 moles of propane will react with oxygen to form x grams of carbon dioxide. Although, there are several ways to calculate it, but I will use my personal way.
I solve this type of problem in two steps. In the first step, I find the number of moles of unknown species like carbon dioxide. I establish relationship of ratio between propane and carbon dioxide.
We can see that it is 1 to 3. 4 moles of propane is given and the moles of carbon dioxide is unknown. I have to cross multiply them. 1 into x is equal to 4 into 3. After calculation, I get x is equal to 12 moles of carbon dioxide. Hence, 4 moles of propane will react with oxygen to form 12 moles of carbon dioxide. Let me repeat it.
4 moles of propane will react with oxygen to form 12 moles of carbon dioxide. In the second step, I just convert the number of moles to grams. The number of moles of carbon dioxide is 12 moles which we calculated. Now the molar mass of carbon dioxide is 44 gram. To convert the number of moles of carbon dioxide to grams, I use this formula.
Number of moles and to molar mass. We know that the number of moles of carbon dioxide is 12 moles and to the molar mass of carbon dioxide is 44 gram. After calculation, I get 528 grams of carbon dioxide. Therefore, we say that 4 moles of propane will react with oxygen to form 528 grams of carbon dioxide.
In such type of problems, I convert given number of moles to moles of unknown species, then I convert number of moles to grams. Hence noted down this important conversion. The third type of stoichiometric conversion is grams to mole conversion.
Consider this problem. How many moles of hydrogen are necessary to react with 6 gram of nitrogen to produce NH3? Well, as usual, I write balanced chemical reaction.
Hydrogen gas plus nitrogen gas will react together to form NH3. I put here 3, 1 and 2. This is the complete balanced chemical equation. To solve these type of stoichiometric problems, I follow two steps. In the first step, I convert the given mass to mole of the known specie. For example, 6 gm of nitrogen gas is given.
and we have to find x moles of hydrogen gas. We know that the given mass of nitrogen gas is 6 g and the molar mass of nitrogen gas is 28 g. Now, I will use this formula to find the number of moles of nitrogen gas, given mass upon molar mass.
The given mass of nitrogen is 6 g and the molar mass of nitrogen gas is 28 g. After calculation, I get 0.2 g. 0.23 moles of nitrogen gas.
Hence 6 gram or 0.23 moles of nitrogen gas react with X mole. In the second step, I will find the number of moles of unknown species like hydrogen gas. I establish relationship of ratio between hydrogen gas and nitrogen gas.
It is 3 to 1. We know that 0.23 moles of nitrogen gas react with X moles of hydrogen gas. I cross multiply them. 1 into x is equal to 3 into 0.23. After calculation, I get x is equal to 0.69 moles of hydrogen gas. Therefore, 6 gram of nitrogen gas will react with 0.69 moles of hydrogen gas to produce NH3.
In such type of problems, I convert given mass to number of moles Then I convert number of moles to number of moles of unknown species. Hence noted down this important conversion. Lastly, let me teach you grams to grams conversion.
For example, consider this problem. How many grams of oxygen react with 10 grams of hydrogen gas to form H2O? Well, as usual, I write the complete balanced chemical equation. Hydrogen gas plus oxygen gas react together to form water. I write 2, 1 and 2. Now it is a complete balanced chemical reaction.
Here the mass of hydrogen gas is given which is 10 grams. I need to find the mass of oxygen gas. Also, I need to find the number of moles of hydrogen gas and oxygen gas. Firstly, I find the molar mass of hydrogen gas.
oxygen gas and H2O. The molar mass of hydrogen gas is 2 gram, that of oxygen gas is 32 gram and that of water is 18 gram. Now I will follow these three steps to solve such type of problems. In the first step, I will find number of moles of non-species like hydrogen gas.
I use this formula. Number of moles of hydrogen gas is equal to given mass upon molar mass. The given mass of hydrogen gas is 10 gm and its molar mass is 2 gm. After calculation, I get 5 moles of hydrogen gas. Thus 10 gm or 5 moles of hydrogen gas react with x gm of oxygen.
In the second step, I find the number of moles of unknown species like oxygen gas. To do so, I establish relationship of ratio between hydrogen gas and oxygen gas. We know that it is 2 to 1. Also we know that 5 moles of hydrogen gas react with x moles of oxygen gas. Now I cross multiply them. 2 into x is equal to 1 into 5. After calculation, I get 2.5 moles.
Hence, 5 moles of hydrogen gas react with 2.5 moles of oxygen gas. Now, I will convert 2.5 moles of oxygen gas to grams. In the third step, I convert the number of moles to grams.
To do so, I use this formula. Mass of oxygen gas is equal to the number of moles and to molar mass. The number of moles of oxygen gas is 2.5 and its molar mass is 32 g. After calculation, I get 80 g of oxygen gas.
Therefore, 10 g of hydrogen gas react with 80 g of oxygen gas to produce H2O. In such type of problems, remember these three steps. Firstly, I convert given mass to number of moles. Secondly, I convert number of moles to number of moles of unknown species.
Thirdly, I convert number of moles to grams. I hope that you have learned all about basic stoichiometry.