Stoichiometry example problem 1 | Physical Processes | MCAT | Khan Academy
Summary
TLDRThe video explains the stoichiometry of the reaction between solid phosphorus and chlorine gas to form phosphorus trichloride. It starts by balancing the chemical equation, ensuring equal atoms on both sides. The process involves converting given grams of phosphorus to moles, determining the required moles of chlorine gas, and calculating the grams needed. Finally, it addresses how to find the amount of phosphorus trichloride produced, emphasizing mass conservation. The explanation includes detailed steps and unit cancellations, making the complex chemistry problem accessible.
Takeaways
- 🧪 The reaction between solid phosphorus and chlorine gas produces phosphorus trichloride.
- 🔄 Stoichiometry is essential for determining the quantities of reactants and products in a chemical reaction.
- ⚖️ The chemical equation must be balanced to ensure the correct stoichiometric ratios.
- 🔍 The atomic weight of phosphorus (P) is approximately 31, and for P4, it's 124 grams per mole.
- 📊 To find the moles of a substance, divide the mass by the molar mass.
- 📚 Moles are a measure of the amount of substance, with one mole containing Avogadro's number of entities.
- 🔄 For every mole of P4, six moles of Cl2 are required according to the balanced chemical equation.
- 📈 The molar mass of Cl2 is approximately 70.906 grams per mole, used to calculate the mass of chlorine needed.
- 📝 The calculation of moles of reactants and products is crucial for understanding the chemical reaction.
- 🧐 Conservation of mass is applied to find the mass of the product, phosphorus trichloride, by adding the masses of the reactants.
- 📉 The total mass of reactants equals the total mass of the products, indicating no mass is lost or gained in the reaction.
Q & A
What is the balanced chemical equation for the reaction between phosphorus and chlorine gas?
-The balanced chemical equation is 4P + 6Cl2 → 4PCl3, representing the reaction of solid phosphorus with chlorine gas to produce phosphorus trichloride.
What is the molar mass of phosphorus (P) in grams per mole?
-The molar mass of phosphorus is approximately 31 grams per mole.
How many grams of a P4 molecule would be equivalent to one mole?
-One mole of P4, which contains four phosphorus atoms, would be 124 grams (4 times the molar mass of a single phosphorus atom).
How many moles of P4 are present in 1.45 grams of solid molecular phosphorus?
-There are 0.0117 moles of P4 in 1.45 grams of solid molecular phosphorus, calculated by dividing the mass by the molar mass of P4.
What is the stoichiometric ratio of chlorine gas to phosphorus in the balanced equation?
-The stoichiometric ratio is 6 moles of chlorine gas to 1 mole of phosphorus.
How many moles of chlorine gas are required to react with 0.0117 moles of P4?
-0.07 moles of chlorine gas are required, based on the stoichiometric ratio from the balanced equation.
What is the molar mass of chlorine gas (Cl2) in grams per mole?
-The molar mass of chlorine gas is approximately 70.906 grams per mole.
How many grams of chlorine gas are needed to react with 1.45 grams of phosphorus?
-4.96 grams of chlorine gas are required, calculated by multiplying the moles of chlorine gas needed (0.07 moles) by its molar mass.
What is the law of conservation of mass, and how does it apply to this chemical reaction?
-The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. In this reaction, the total mass of reactants (phosphorus and chlorine gas) equals the mass of the product (phosphorus trichloride).
How can you calculate the mass of phosphorus trichloride produced in the reaction?
-You can either use the stoichiometry of the reaction to find the moles of PCl3 produced and then convert to mass, or apply the law of conservation of mass, which states the mass of the product will be the sum of the masses of the reactants.
What is the significance of balancing a chemical equation before performing stoichiometry calculations?
-Balancing a chemical equation ensures that the number of atoms of each element is the same on both sides of the equation, which is necessary for accurate stoichiometry calculations and reflects the actual chemical reaction taking place.
Outlines
🧪 Stoichiometry and Balancing Chemical Equations
The script begins with an introduction to a chemistry problem involving the reaction of solid phosphorus with chlorine gas to produce phosphorus trichloride. The voiceover emphasizes the importance of balancing the chemical equation before proceeding with stoichiometry calculations. The equation is presented with the reactants and products, and the process of balancing the equation by ensuring an equal number of phosphorus and chlorine atoms on both sides is explained. The molecular weights of phosphorus and the resulting phosphorus trichloride are discussed to facilitate the calculation of moles from given masses.
📚 Calculating Moles and Reactant Requirements
This paragraph delves into the calculation of moles of phosphorus from the given mass, using the atomic weight of phosphorus to determine the mass of one mole of molecular phosphorus. The script performs a dimensional analysis to convert grams of phosphorus to moles. It then uses the balanced chemical equation to determine the stoichiometric ratio between phosphorus and chlorine gas, calculating the moles of chlorine gas required to react completely with the given amount of phosphorus. The atomic weight of chlorine is introduced to convert moles of chlorine gas into grams, answering the first part of the query regarding reactant requirements.
🔍 Mass Conservation and Product Calculation
The final paragraph addresses the calculation of the mass of phosphorus trichloride produced from the reaction. Two methods are mentioned: a straightforward mass conservation approach and a more detailed molar mass calculation. The script opts for the mass conservation method, which states that the total mass of reactants equals the total mass of products. By adding the masses of the reactants (phosphorus and chlorine gas), the script concludes with the total mass of phosphorus trichloride produced, thus answering the second part of the query regarding product formation.
Mindmap
Keywords
💡Phosphorus
💡Chlorine Gas
💡Phosphorus Trichloride
💡Stoichiometry
💡Balanced Equation
💡Moles
💡Atomic Weight
💡Molecular Weight
💡Dimensional Analysis
💡Law of Conservation of Mass
💡Mole Ratio
Highlights
Solid phosphorus reacts with chlorine gas to produce phosphorus trichloride.
The stoichiometry problem involves calculating reactants and products in a chemical reaction.
The chemical equation must be balanced before proceeding with stoichiometry calculations.
Balancing the equation involves adjusting coefficients to equalize the number of atoms on both sides.
Moles are used to relate the amount of reactants and products in a chemical reaction.
The atomic weight of phosphorus is approximately 31, which is crucial for mole calculations.
Molecular phosphorus (P4) has an atomic weight of 124, derived from the atomic weight of single phosphorus atoms.
1.45 grams of molecular phosphorus is converted into moles using dimensional analysis.
The stoichiometric ratio between phosphorus and chlorine gas is one to six moles.
0.0117 moles of molecular phosphorus require 0.07 moles of chlorine gas.
The atomic weight of chlorine is approximately 35.453, and molecular chlorine (Cl2) is double that.
4.96 grams of chlorine gas are required to react with 1.45 grams of phosphorus.
Conservation of mass principle states that the total mass of reactants equals the total mass of products.
An alternative method to find the mass of phosphorus trichloride produced is through stoichiometric ratios.
The easy way to determine the mass of the product is by using the conservation of mass.
The total mass of reactants (1.45g P + 4.96g Cl2) equals the mass of the product formed.
Transcripts
- [Voiceover] We know that solid phosphorus
will react with chlorine gas
to spontaneously, to produce
phosphorus trichloride,
liquid phosphorus trichloride,
and we're told that we have 1.45 grams
of solid molecular phosphorus and we're asked,
"How many grams of chlorine is required
"to essentially use up all of the phosphorus that we have,
"and how many grams of phosphorus trichloride
"is going to be produced?"
Now before you do any of these stoichiometry problems,
and that's just a fancy word
for problems where you need to figure out
how much of a certain reactant is required,
or how much of a product is going to be produced.
Before you do any of these problems,
you have to be sure that your reaction,
or that your equation is balanced.
So let's make sure.
So on the left-hand side here,
this molecule of phosphorus has four phosphorus atoms,
so on the whole left-hand side are,
all of our reactants combined have four phosphorus atoms.
So our products need to also have four phosphorus atoms,
but the way it's written right now, we only have one.
So let me just multiply this guy
by four.
Now I have four phosphorus atoms
on both sides of the equation.
Let's balance the chlorine now.
On the left-hand side I only have two chlorine atoms.
This one molecule of chlorine has two atoms in it.
Here I have, each molecule of phosphorus trichloride
has three chlorines, and I have four molecules of it,
so four times three, I have 12 chlorines
on the right-hand side.
So I have 12 on the right-hand side,
I need to have 12 on the left, I only have two here.
Let me multiply that by six.
Six times two is 12,
four times three is 12.
Now our equation is all balanced.
Four phosphoruses on each side,
and 12 chlorines.
Now the next thing we have to do,
now that we know that we have a balanced equation,
and we can kind of get into the meat of the problem,
is figure out how many moles of phosphorus
we're dealing with.
Because once we know the moles,
we can use the stoichiometric ratios,
which is just essentially saying, "Look,
"for every mole of that, I need six moles of that,
"and for every mole of that,
"I'm gonna produce four moles of that."
So you want to get it all in terms of moles.
So let's figure out how many moles of phosphorus
we have on our hands.
And let's look at our periodic table.
This periodic table, I have to give
proper attribution to the maker.
It's made by Levon Han Cedric,
I got this off of Wikimedia Creative Commons,
it had an attribution license.
Other than that, we're free to use it.
And lets go to phosphorus.
Phosphorus right here,
phosphorus right here has its atomic weight
of 30.974.
Let's just round that up to 31.
Let me write it down right here.
So phosphorus,
phosphorus has an atomic weight
atomic weight,
of 31,
which tells us that a mole of phosphorus
will weigh 31 grams.
Remember a mole is this, you know,
6.02 times 10 to the 20th,
this huge number of atoms.
If you have that many number of atoms
of atomic phosphorus,
it's going to weigh 31 grams.
Now if you look at the atomic weight of P four,
or a molecule that has four phosphorus atoms in it
it's gonna be four times this.
So it's going to have an atomic
atomic weight of,
well what's four times 31?
It's 124,
of 124,
which means that one mole of,
let me write it here.
This tells us right there,
that one mole
of solid molecular phosphorus
is going to have a well,
since we're doing grams, I should say,
will have a mass of
124 grams.
Now, given that,
we can use that with this information
to figure out how many moles
of molecular phosphorus we have.
Solid molecular phosphorus.
So let me start over here.
So we have 1.45 grams,
let me write it this way,
grams of phosphorus,
of molecular phosphorus,
each molecule containing four atoms,
and we can just do a little dimensional analysis,
make sure everything cancels out,
times this information right here.
We have one mole of
phosphorus, of molecular phosphorus
for every 124
grams of molecular phosphorus.
I should write this here,
just so we remember what we're talking about.
120, one mole of molecular phos
for every 124 grams
of molecular phosphorus.
And then this cancels out.
So what we have, the units at least cancel out.
The grams of phosphorus, grams of phosphorus,
and then we get 1.45
times, what do we have,
times one over 24,
times one over 124 moles,
moles of molecular phosphorus.
And we could figure that out.
Is equal to 0.0
1-1-7 moles of phosphorus.
That's what we're starting off with.
That' s what this 1.45 grams are.
And that makes sense.
If we had an entire mole of molecular phosphorus
it would weigh 124 grams.
We only have 1.45 of that,
so it's almost, you know,
it's a little bit more than one hundredth,
which makes sense.
This number right here is a little more
than one hundredth.
Now we need to think about for every mole,
let's scroll down here, although I don't want to lose
my equation.
We need to think about for every mole,
so let me write down what we just,
so we have 0.0117
moles of molecular phosphorus,
and for every mole of phosphorus,
how many moles of chlorine molecules do we need?
So let's, let me write this down.
So for every, and I'll write it this way.
For every six moles of chlorine gas,
I'll do it in blue.
So times,
for every six moles
of chlorine gas, we need
one mole, we need one mole
of molecular phosphorus.
And the reason I wrote it this way
instead of writing one mole of molecular phosphorus
for every six moles of chlorine gas,
is because I want to make sure
that the units cancel out.
And it also should make sense for you intuitively.
If I have, if I have one of this,
I'm gonna need six of this.
If I have
0.0117 of this,
I'm gonna need six times of the chlorine gas,
and this, the units work out.
This cancels out with that,
and I'm just essentially multiplying by six.
So we're going to have six times,
let me multiply it,
six time 0.0117,
so this is equal to,
0.07 moles of
chlorine gas required.
And I could write it right here.
Let me write "required."
Required in the numerator, required in the denominator.
That makes a little bit more sense or it clarifies things.
For every six moles of chlorine gas that are required,
one mole of phosphorus,
of solid phosphorus is required.
Or you could say for every mole of phosphorus is required,
you need six moles of chlorine gas.
We got that just from the original equation.
So I'll write this "required" down right there.
So we're almost there.
We figured out that 0.07 moles
of chlorine gas are required.
But what we want to find out is how many grams
of chlorine gas are required,
so we just have to figure out how many grams
there are per mole.
So let's figure that out.
Let's look at chlorine.
Chlorine is off here to the right.
This is a super huge periodic table.
Chlorine right here has an atomic weight
of 35.453.
So chlorine, chlorine,
35.453,
atomic weight.
The weighted average of all of the isotopes
of chlorine on earth.
So chlorine, molecular chlorine gas,
which has two atoms, is going to have twice that,
so I could do it in my head,
it's gonna be right about what,
70, 70.906,
atomic weight,
which tells us that,
well let me write this information that we had down before,
we are required to have
.07 moles of chlorine gas,
that is what is required,
and let's multiply it.
We want moles in the denominator.
So we want moles in the denominator here,
so one mole
of chlorine,
of chlorine gas,
for every mole, how much is that?
How much is that, what's the mass going to be?
Well if the atomic weight of chlorine gas
is 70.906, that means
one mole of it is going to have a mass of
70.906 grams.
So for every mole, we have 70.90 grams,
we have .07 moles,
so we'll multiply .07 times 70
to figure out how many grams we have.
And the units cancel out.
We have moles of chlorine gas,
moles of chlorine gas,
and we're just gonna have grams required.
So this is going to be equal to 70
times .07, or
.96 grams of,
and actually, this is grams of CL,
of chlorine gas, I should write that there, so
grams of chlorine gas are
required, even the required worked out
with the dimension analysis,
and we've answered the first part of the problem.
If we have 1.45
grams of phosphorus
as one of the reactants,
then we're gonna need 4.96,
4.96 grams of chlorine are required.
Now the second question is how much
phosphorus trichloride is produced.
Now there's two ways to do it,
an easy way and a hard way.
The easy way says, "Well, mass is conserved."
You know, if you have grams on this side,
the total mass on this side has to be
equal to the total mass on that side.
That's the easy way.
The slightly harder way is,
you could have done the exact same thing
we did to the chlorine gas,
you could do it with the phosphorus trichloride.
You could say, "Hey, for every mole of this,
"I need four moles of that,"
and then figure out the mass of each mole
and multiply and do all of that.
But let's do it the easy way.
We know that we have
1.45 grams of this reactant,
and we have 4.96 grams of this reactant,
so this, which is the only product right here,
this product right here,
that must have a mass of our combined reactants,
'cause it's going to react fully.
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