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Stoichiometry | Mole to mole | Grams to grams | Mole to grams | Grams to mole | Mole ratio — Transcript

by Najam Academy · 2,279 words · 326 segments · language en · Watch on YouTube

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0:00basic introduction to Stoichiometry I
0:03will teach you all the important
0:04stoichiometric conversions using Matrix
0:08firstly let me teach you that what is
0:11stoichiometry well the word strike you
0:14means element and the word metri means
0:17measurement so Stoichiometry is the
0:20calculation of products and reactants
0:22and a chemical reaction for example
0:26consider this general chemical reaction
0:28with the help of Stoichiometry we can
0:31find the number of moles of reactants
0:34and the number of moles of products so
0:37remember that Stoichiometry is the
0:40calculation of products and reactants
0:42and a chemical reaction now let me teach
0:46you the important concept of
0:47coefficients and a chemical reaction
0:49which a lot of students are not
0:51understanding for example consider this
0:55chemical reaction hydrogen gas react
0:57with nitrogen gas to form a ammonia NH3
1:01now I will balance this chemical
1:03reaction there are two hydrogen atoms in
1:06the reactants well there are three
1:09hydrogen atoms in the products I place
1:12two with products and three with the
1:14reactants now three n two two is equal
1:17to 6 atoms 2 into 3 is equal to 6 atoms
1:22there are two nitrogen atoms and there
1:25are also two nitrogen atoms so this is
1:29complete balanced chemical reaction now
1:32these three with hydrogen gas is known
1:34as coefficient of hydrogen gas
1:37secondly this one with nitrogen gas is
1:40known as coefficient of nitrogen gas
1:43while these two with NH3 is known as
1:46coefficient of NH3 here I will teach you
1:50three different stories of coefficients
1:53firstly the coefficients represents the
1:56amount of substances for example this
2:00three coefficient of hydrogen gas shows
2:023 moles of hydrogen gas
2:05one coefficient of nitrogen gas
2:07represents one mole of nitrogen gas and
2:10two coefficient of NH3 represents 2
2:13moles of NH3 secondly coefficients
2:17represent a ratio of reactants to
2:19products for example I write hydrogen
2:23gas ratio to nitrogen gas to NH3 we know
2:27that the coefficient of hydrogen gas is
2:303 that of nitrogen gas is 1 and that of
2:33NH3 is 2. hence this is the ratio of
2:37reactants to products thirdly
2:40coefficients represents number of
2:42molecules let I write three hydrogen gas
2:46one nitrogen gas two NH3 now this three
2:50coefficient of hydrogen gas represents
2:52three molecules of hydrogen gas this one
2:56coefficient of nitrogen gas represents
2:59one molecule of nitrogen gas and these
3:03two coefficient of NH3
3:05represents two molecules of NH3 just
3:09remember that Story 1 tells that the
3:12coefficient represents the number of
3:14moles Story 2 tells that the coefficient
3:18represents the ratio of reactants to the
3:20products Story 3 tells that the
3:24coefficient represents the number of
3:25molecules
3:27now how can we use these stories to
3:30calculate the amount of reactants and
3:32products
3:33well consider this question how can you
3:37form six molecules of NH3 the answer is
3:40simple according to the Second Story the
3:43ratio of hydrogen to nitrogen to NH3 is
3:46three to one one to two we need 6
3:50molecules of NH3 I just multiply 2 N to
3:543 I also multiply the ratio of n 2 by 3
3:58and that of H2 by 3 now 2 and 2 3 is
4:03equal to 6 1 into 3 is equal to 3 3 into
4:073 is equal to 9. this reveals that we
4:11need 9 molecules of hydrogen gas to
4:14react with three molecules of nitrogen
4:16gas in order to form six molecule of NH3
4:21let me repeat this important statement
4:23we need 9 molecules of hydrogen gas to
4:27react with three molecules of nitrogen
4:30gas in order to form six molecules of
4:33NH3 to conclude this whole concept any
4:37coefficient and a reaction shows either
4:40number of moles a ratio or number of
4:44molecules and snorted down all these
4:47important Concepts
4:49now let me teach you the different
4:51conversions of Stoichiometry like mold
4:54to mole conversion consider this problem
4:56how many moles of nitrogen gas is needed
5:00to react with 13.5 moles of hydrogen to
5:03form NH3 well I Write the balanced
5:07chemical reaction we know that hydrogen
5:10gas plus nitrogen gas react together to
5:13form NH3 I put here 3 1 and 2. this is
5:18the complete balanced chemical reaction
5:20now according to the given statement
5:2313.5 moles of hydrogen gas is given we
5:27need to find the number of moles of
5:29nitrogen gas although there are several
5:32ways to calculate it but I will teach
5:35you my way to calculate such type of
5:38questions
5:39firstly I write ratio of hydrogen to the
5:42ratio of nitrogen gas we know that it is
5:46three to one now listen carefully we are
5:49already given the number of moles of
5:51hydrogen I will write this 13.5 moles
5:55below the hydrogen gas let me repeat it
5:59we are already given the number of moles
6:02of hydrogen gas I will write this 13.5
6:05moles below this hydrogen gas
6:09here if three moles of hydrogen gas
6:12react with one mole then 13.5 moles of
6:15hydrogen gas react with X moles now I
6:19will just cross multiply them 3 into X
6:22is equal to 1 into 13.5
6:27I divide both sides by 3 after
6:30calculation I get X is equal to 4.5
6:34moles this we need 4.5 moles of nitrogen
6:38to react with 13.5 moles of hydrogen
6:42hence noted down this important
6:44conversion
6:45secondly consider this problem how many
6:49moles of sulfur trioxide will form when
6:518.5 moles of sulfur dioxide react with
6:55oxygen firstly I Write the balanced
6:58chemical reaction according to the given
7:01statement sulfur dioxide plus oxygen gas
7:05react together to form sulfur trioxide
7:08now two oxygen atoms plus two oxygen
7:12atoms is equal to four oxygen atoms
7:15while in the products there are three
7:18oxygen atoms I place here too and the
7:22product there are two sulfur atoms in
7:25the reactant I plus here too hence this
7:29is the complete balanced chemical
7:31equation according to the given
7:34statement
7:358.4 moles of sulfur dioxide will react
7:38with oxygen to form X moles of sulfur
7:41trioxide now I will use my personal way
7:45to calculate the number of moles of
7:47sulfur trioxide I established ratio
7:51between sulfur dioxide and sulfur
7:53trioxide we can see that it is 2 ratio
7:56to 2 the number of moles of sulfur
7:59dioxide is 8.4 moles we need to
8:03calculate the number of moles of sulfur
8:05trioxide let it is X now I cross
8:09multiply them 2 into X is equal to 2
8:13into 8.4 I divide both sides by 2 after
8:17calculation I get 8.4 moles thus 8.4
8:23moles of sulfur dioxide will react with
8:26excess of oxygen to form 8.4 moles of
8:30sulfur trioxide hence noted down this
8:33important problem
8:35the second type of stoichiometric
8:37conversion is mole to gram conversion
8:40for example consider this problem
8:43firstly I write the complete balanced
8:47chemical reaction according to the given
8:49statement propane c3h8 react with oxygen
8:54to form carbon dioxide plus water it is
8:58a simple combustion reaction now there
9:01are three carbons in the reactants and
9:04one carbon and the products I plus here
9:073 there are 8 hydrogen in the reactants
9:10and two hydrogen in the products I place
9:14here for 2 into 4 is equal to 8 hydrogen
9:18atoms now 2 and 2 3 is equal to 6 oxygen
9:22atoms plus four oxygen atoms is equal to
9:2610 oxygen atoms I write here 5 hence
9:30this is the complete balanced chemical
9:32equation now four moles of propane will
9:35react with oxygen to form X grams of
9:39carbon dioxide
9:41Alto there are several ways to calculate
9:44it but I will use my personal way I
9:47solve this type of problem in two steps
9:50in the first step I find the number of
9:53moles of unknown species like carbon
9:56dioxide I establish relationship of
9:59ratio between propane and carbon dioxide
10:01we can see that it is one two three four
10:05moles of propane is given and the moles
10:08of carbon dioxide is unknown I have to
10:12cross multiply them 1 into X is equal to
10:154 into 3 after calculation I get X is
10:21equal to 12 moles of carbon dioxide
10:24hence four moles of propane will react
10:26with oxygen to form 12 moles of carbon
10:30dioxide let me repeat it four moles of
10:33propane will react with oxygen to form
10:3612 moles of carbon dioxide
10:39and the Second Step I just convert the
10:42number of moles to grams the number of
10:45moles of carbon dioxide is 12 moles
10:48which we calculated now the molar mass
10:51of carbon dioxide is 44 gram to convert
10:55the number of moles of carbon dioxide to
10:57grams I use this formula
11:00number of moles into molar mass we know
11:03that the number of moles of carbon
11:05dioxide is 12 moles and two the molar
11:09mass of carbon dioxide is 44 gram after
11:12calculation I get
11:15528 grams of carbon dioxide therefore we
11:21say that 4 moles of propane will react
11:23with oxygen to form 528 grams of carbon
11:28dioxide in such type of problems I
11:32convert given number of moles to moles
11:34of unknown species then I convert number
11:37of moles to grams hence noted down this
11:40important conversion the third type of
11:43stoichiometric conversion is grams to
11:45mole conversion consider this problem
11:48how many moles of hydrogen are necessary
11:51to react with 6 gram of nitrogen to
11:55produce NH3 well as usual I write
11:59balanced chemical chemical reaction
12:01hydrogen gas plus nitrogen gas will
12:04react together to form NH3 I put here 3
12:081 and 2. this is the complete balanced
12:12chemical equation
12:13to solve this type of stoichiometric
12:16problems I follow two steps in the first
12:19step I convert the given Mass to mole of
12:22the known species for example 6 gram of
12:25nitrogen gas is given and we have to
12:28find X moles of hydrogen gas we know
12:31that the given mass of nitrogen gas is 6
12:34gram and the molar mass of nitrogen gas
12:37is 28 gram now I will use this formula
12:40to find the number of moles of nitrogen
12:43gas given Mass upon molar mass the given
12:47mass of nitrogen is 6 gram and the molar
12:50mass of nitrogen gas is 28 gram after
12:53calculation I get
12:560.23 moles of nitrogen gas and 6 gram
13:00are 0.23 moles of nitrogen gas react
13:03with X mole and the Second Step I will
13:07find the number of moles of unknown
13:08species like hydrogen gas I establish
13:12relationship of ratio 2 between hydrogen
13:15gas and nitrogen gas it is three to one
13:19we know that 0.23 moles of nitrogen gas
13:23react with X moles of hydrogen gas I
13:26cross multiply them 1 and 2x is equal to
13:313 and 2 0.23 after calculation I get X
13:37is equal to 0.69 moles of hydrogen gas
13:41therefore 6 gram of nitrogen gas will
13:45react with
13:460.69 moles of hydrogen gas to produce
13:50NH3 in such type of problems I convert
13:54given Mass to number of moles then I
13:57convert number of moles to number of
13:58moles of unknown species hence noted
14:02down this important conversion lastly
14:05let me teach you grams to grams
14:07conversion for example consider this
14:10problem how many grams of oxygen react
14:14with 10 grams of hydrogen gas to form
14:17H2O well as usual I write the complete
14:21balanced chemical equation hydrogen gas
14:24plus oxygen gas react together to form
14:27water I write 2 1 and 2. now it is a
14:31complete balanced chemical reaction here
14:35the mass of hydrogen gas is given which
14:38is 10 grams I need to find the mass of
14:41oxygen gas also I need to find the
14:44number of moles of hydrogen gas and
14:46oxygen gas firstly I find the molar mass
14:50of hydrogen gas oxygen gas and H2O the
14:54molar mass of hydrogen gas is 2 gram
14:57that of oxygen gas is 32 gram and that
15:00of water is 18 Gram now I will follow
15:03these three steps to solve such type of
15:06problems in the first step I will find
15:09number of moles of non-specy like
15:12hydrogen gas I use this formula number
15:16of moles of hydrogen gas is equal to
15:18given Mass upon molar mass the given
15:22mass of hydrogen gas is 10 gram and its
15:25molar mass is 2 gram after calculation I
15:29get 5 moles of hydrogen gas thus 10 gram
15:32are 5 moles of hydrogen gas react with X
15:35grams of oxygen and the Second Step I
15:39find the number of moles of unknown
15:40species like oxygen gas
15:43to do so I establish relationship of
15:46ratio between hydrogen gas and oxygen
15:49gas we know that it is two to one also
15:54we know that 5 moles of hydrogen gas
15:56react with X moles of oxygen gas now I
16:00cross multiply them 2 into X is equal to
16:031 into 5. after calculation I get 2.5
16:08moles hence 5 moles of hydrogen gas
16:11react with 2.5 moles of oxygen gas now I
16:15will convert 2.5 moles of oxygen gas to
16:19grams and the third step I convert the
16:22number of moles to grams to do so I use
16:25this formula mass of oxygen gas is equal
16:29to number of moles and to molar mass the
16:33number of moles of oxygen gas is 2.5 and
16:37its molar mass is 32 grams after
16:40calculation I get 80 grams of oxygen in
16:44case therefore 10 grams of hydrogen gas
16:47react with 80 grams of oxygen gas to
16:51produce H2O in such type of problems
16:54remember these three steps firstly I
16:58convert given Mass to number of moles
17:00secondly I convert number of moles to
17:04number of moles of unknown species
17:05thirdly I convert number of moles to
17:09grams I hope that you have learned all
17:12about basic stoichiometry

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