KRISTEN CACCIATORE: Welcome to AP Daily Practice sessions.
I'm so glad you're here.
Today in session 4, we're doing multiple-choice questions
from units 4, 5, and 6.
And those are going to appear on the exam.
So let's get you ready.
Before we launch into our first problem,
if you want to work through these problems on your own,
you can click on the link you see
in the video now or in the YouTube description
and download a PDF of the problems.
All right.
First one, so Fe solid plus 2HCl aq
produces FeCl2 s plus H2 gas.
When a student adds 30 milliliters of 1.00 molar
hydrochloric acid to 0.56 grams of powdered iron,
a reaction occurs according to the equation above.
When the reaction is complete at 273 Kelvin and 1.0 atmospheres,
which of the following is true?
All right.
So we've got HCl is in excess, and 0.100 mole of H2 HCl
remains unreacted.
HCl is in excess, and 0.020 moles of HCl remains unreacted.
0.015 moles of FeCl2 has been produced.
Or 0.22 liters of H2 has been produced.
Well, this is clearly a stoichiometry problem
because the answer choices are all
asking us about the amount of reactants and products present.
So let's do some stoichiometry.
So we've got our balanced chemical equation here,
and we can see we have a 1 to 2 ratio of the reactants
to each other.
We also have amounts of reactants present.
So we can figure out which one is the limiting reactant.
And you'll notice that the mole ratio requires
twice as much HCl as Fe, but the system
contains three times as much HCl, 0.03, moles as iron,
0.01 moles, which means that HCl is in excess.
So we've figured that out, which is great.
The start of the stoichiometry problem
is knowing the limiting and excess reagents.
That doesn't really help us eliminate any answer choices
at this point, but we're going to get there.
Now, let's think about how much product
is produced based on the amount of reactants
and the limiting reactant.
So all of the limiting reactant iron is going to react
and twice as much HCl, 0.02 moles is going to react.
And that's going to make 0.01 moles of each product using
the stoichiometry, the mole ratios,
and leave us 0.01 moles of HCl unreacted.
So based on that, we can eliminate our choice A
because it says that we have 0.100 mole of HCl unreacted,
and that's not the case.
Likewise for B has the wrong amount of HCl unreacted.
C is the incorrect amount of unreacted chloride produced
because we know that it's 0.01 mole.
And that leaves us with D. And D is in liters.
And remember that this is at STP 273 Kelvin and 1 atmosphere,
so 22.4 liters is 1 mole.
And 0.01 moles is 0.22 liters of H2.
And that is choice D. So D is right.
That was a lot of work for one point, but we did it.
All right, next question.
So we have these two steps in a chemical reaction mechanism.
It says, which of the following represents the overall chemical
equation and the rate law for the reaction?
All right.
So let's take this one part at a time.
We have four answer choices, and they each
have a chemical equation in a rate law.
So I'm going to zero in on the overall chemical equation
first and take a look at what I've got here.
Notice that I can cancel out the HI in step 1
and step 2 because they appear on opposite sides
of the equation.
And then when I add them together,
I get the overall equation H2 gas plus 2ICl gas
produces 2HCl gas plus I2 gas.
All right.
So then I want to check the answer choices
and see if I can eliminate any of them.
Well, notice that in answer choice A,
it has the wrong amount of HCl.
So I can cross that out right away.
The other choices all look like they
have the right overall chemical equation.
So I can't eliminate any more yet.
So now, let's go ahead and zero in on the rate
law, which is the second thing that the problem asks us for.
Well, we need to focus on the slow step
because that determines the rate law.
And we look at the reactants for the slow step
and the rate law is just going to be proportional to reactants
in the slow step raised to the power
of their stoichiometric coefficients,
which in this problem is 1 for both H2 and ICl.
So then we just need to find which of the answer
choices has that rate law, and we can see here
that its answer choice C has the two reactants raised
to the first power.
And so C is our correct choice.
All right.
Problem number three.
We have a particle diagram.
The diagrams above represent solutes
present in two different aqueous solutions
before they were mixed.
Water molecules are not shown.
When the solutions were combined,
a precipitation reaction took place.
Which of the diagrams below is the best particle
representation of the mixture after the precipitation
reaction occurred?
All right.
So we've got four choices, four particle diagrams.
First thing we need to recognize is
that we've got a precipitate.
So we have a solid forming here.
What is that solid?
Well, we know that both sodium and nitrate ions are always
soluble.
That's something you have to know.
Group 1 metals and nitrate ions always soluble in water.
So therefore, the precipitate has
to form from the other two ions in the system,
and that is silver ions and chloride ions.
So we get AgCl.
That means that in our diagram, we
need to see AgCl at the bottom of the container.
Because it's a solid, it has precipitated.
And so we need to X out A because the AgCl is
at the bottom and we can also X out C because the AgCl is not
at the bottom.
In both B and D, we see the AgCl correctly
represented-- so we need to move on to something else.
Well, now let's focus on the spectator ions--
the sodium and the nitrate ions.
They should be shown as separated
because ions in solution are physically
separated from each other.
They're solvated by water molecules.
And so that means that we can X out choice B,
because it has those two ions connected to each other.
And that leaves us with D as our correct choice.
All right.
Problem number 4.
All right.
This looks like some lab data here.
A student places a sample of a pure metal in a crucible
and heats it strongly in air.
Data from the experiment are given in the table above.
The final mass was determined after the sample
was cooled to room temperature.
Which of the following statements related
to the experiment is correct?
All right.
Well, we've got-- the mass of the sample
decreased, so physical changes occurred as the metal first
melted and then boiled out of the crucible;
the mass of the sample increased,
so which chemical change occurred when bonds formed
between the metal and another substance;
there was nothing for the metal to react with,
so only a physical change could have occurred;
or the sample was only heated, so neither a physical
nor a chemical change occurred.
All right.
Well, first we got to look at the data.
And we can see here that we've got a pure metal heated
strongly in air.
Well, air contains molecules of oxygen and nitrogen.
And oxygen and metals react.
So it's possible that there was a chemical reaction
in this situation.
Also notice that the mass increased.
So that is further indication of a chemical change.
And so that tells us that choice B is correct because we
had metal reacting with air.
A to be wrong because it said decreased.
C says there's nothing for the metal to react with.
That's not true.
There's air, which has oxygen and also nitrogen,
but nitrogen doesn't react very much.
And then D says the sample was only heated.
That's not true.
We know that chemical changes can be instigated
by heating in some situations.
All right.
And how about number five?
We have this chemical equation MgCl2 aqueous
plus 2NaOH aqueous produces NaCl aqueous plus MgOH2 solid.
So we're making a precipitate again.
100 milliliters sample of 0.1 molar magnesium chloride
and 100 milliliters sample of 0.2 molar sodium hydroxide
were combined and magnesium hydroxide precipitated
as shown by the equation above.
If the experiment is repeated using solutions
of the same molarity, which of the following changes in volume
will double the amount of magnesium hydroxide produced?
All right.
So we need to think about the amounts of reactants present,
and then if we want to double the amount of the solid
that we make, what do we need to do?
Well, first of all, I need you to notice that we have a 2 to 1
mole ratio of sodium hydroxide to magnesium
chloride in the system because we have same volume,
but the molarity of the sodium hydroxide is double.
And we also have that exact same ratio in our chemical equation.
That tells us that we have exactly the right amount, what
we call a stoichiometric ratio of the two reactants
present in the system as it exists.
That means that both reactants are reacting completely.
They're both limiting, essentially.
So if we want to double the amount of product that is made,
we need to double both of those reactants
because we need all of them in this ratio.
And so that's why choice D is going to be correct.
Because we have the correct ratio to begin with,
so we need to double if we want to double the amount of product
produced.
All right.
Last one for today.
Which of the following combinations
represents the individual reactions and the quantities
needed to determine the heat of reaction delta H0
for the overall reaction represented by the chemical
equation above?
So we have four answer choices here.
Each of them is it's possible to add those two
equations together, each of the answer choices,
to get an overall equation.
We want to pick the answer choice
where if we add the two equations together, then
we're going to end up with that target equation--
CS2 liquid plus 2H20 liquid produces CO2 gas plus 2H2S gas.
So that's what we want to look for.
How do you go about doing that?
Well, what I want you to do is focus on the reactants first,
and notice that we need 1CS2 and 2H20 on the reactant side
of our overall equation.
And so when I go through these answer choices,
I can eliminate answer choices where that is not the case.
Notice in B, I have CS2 on the reactant side and I have H20,
but it's only one.
I need two in my target equation.
So B is out.
Let's look at C. Well, I've got CS2 in the reactant side,
but I don't have H2 on the reactant side.
It's on the product side in C. So C is out.
How about D?
Well, D, I've got CS2, again, on the reactant side,
but again, I have the water on the product side.
So D is out.
And if I go and look at A-- well,
A has the correct reactants.
If I then go and add the two equations together
to double-check that I get the whole correct equation,
I'll see that A is correct.
So there you go.
All right.
Well, that's going to be all for this video,
but we have so many more resources for you
to help you prepare for the exam.
There's AP Daily with videos on every topic in units 4
and 5 and 6.
We also have lots of practice sessions like this one.
There's seven more.
And we have practice sessions from the last two
years on YouTube.
So I encourage you to check those out and best of luck
studying for the exam.
Thank you so much for watching, and I'll catch you next time.