We’re focusing on an example of solution

stoichiometry. So our example states, according to the label on a bottle of

concentrated hydrochloric acid, the contents are 36% HCl by mass and have a

density of 1.18 grams per milliliter. We’re asked three different

questions here ,our first one is, “What is the molarity of the concentrated HCl?” So

let’s start off with that one. Identify what we’re given and what we’re finding

as always, so we’re given 36% of HCl by mass. Now whenever you see this

percentage by mass, that’s gonna act as your conversion factor. So what we’re

gonna do is we’re gonna split this apart and 36 grams of HCl will go on

top and then on the bottom would be a 100 grams of HCl solution. Next we’re

also given our density of 1.18 grams per milliliter. And this is

specifically of the HCl solution. Next we’re asked to find the molarity and

remember, molarity is mole’s divided by liters. So some way somehow we’re gonna

get to moles of HCl and divide that by the liters of our HCl solution. So what

we’re gonna do first is, taking that 36 grams per every 100 gram of

HCl solution, we’re gonna put that and multiply that by the density that was

provided because our grams of HCl and grams of HCl solution are going to

then cancel and then now we have grams of HC divided by the milliliters of our

HCl solution. And then so we’re at this current step right now. After we’ve

multiplied already we have to keep going because remember we’re trying to find

the molarity, so that means we want our moles of HCl on top divided by liters of

our HCl solution. So taking what we just found that 0.42

for 8 grams of HCl, then I’m gonna change this to moles using the molar mass of

HCl. So aligning our grams of HCl and remember grams of HCl how we find that, how

we found the molar mass is just taking our mass of each individual element, we

have hydrogen that has 1.01 chlorine which has 35.45 and we’re going to add those two together to get our molar mass of

HCl which would be 36.46 grams per mole. So I’m going to put

the grams on the bottom and the moles on top since we’re trying to find moles on

top. And then from here our grams of HCl and grams of HCl will then cancel and

then now we have moles of HCl on top and milliliters on the bottom. So we’re

almost done. The next portion here is to change milliliters on the bottom to

liters. So what we’re going to use is 1,000 milliliters, put that on top so our

units would then cancel and we always want those units across from each other

so they can cancel. And then next our liters of HCl is finally on the bottom.

So what we’re going to see is we’re finally going to have our unit of moles

of HCl divided by the leaders of our HCl solution. And then from there

our final answer would give us 11.7 molarity of our HCl solution,

so that’s the molarity of HCl. Let’s move on to Part B. Part B states, “What volume

of it would you need to prepare 283 milliliters of a 2.20 molar

solution of HCl so we’re gonna use our previous portion, our previous molarity

in Part A that we found which was that 11.7 molarity of HCl, so

that is gonna be our given, one of our givens in this question. Now we’re asked

to find another volume, so what I want you to notice is the fact that we

started offer in Part A we had 11.7 molarity of HCl but this

time we decreased our molarity, so anytime we’re decreasing that molarity

and we’re dealing with molarity and volume that’s a dilution. So we have to

use our dilution formula of M1V1 those two are multiplied and set that

equal to M2 x V2. Let’s identify what our M 1 and V 1 and M 2 and V 2 are. So

our M 1 is going to be what we found initially, what we found in Part A, and

then what we’re given our molarity to is that 2.20 molarity of HCl. Next were

asked what volume of it, meaning what volume of our previous molarity, so our

initial molarity. So we’re finding V 1 and they give us V 2 is 283 milliliters.

Okay now that we’ve identified everything here, let’s just plug in. So

we’ll plug in 11.7 for our M 1 and then V 1 remember that’s

what we’re solving for and then M 2 we’ll plug in 2.20 and V 2

we’ll plug in 283 and then from there all we have to do is to get V 1 by itself

so we’ll divide over that eleven point seven to either side and then from there

these two would cancel we’ll solve for V 1 after we do the math so multiply the

numerator, first what’s on top and divide it by 11.7, we would get

53.2 milliliters of HCl. One more question let’s go to Part C.

Part C states, “What mass of sodium bicarbonate would be

needed to neutralize the spill if a bottle containing 1.75 liters of

concentrated HCl dropped on a lab floor and broke open? So once again let’s

identify what are we given, what are we finding? So we’re given that 1.75 liters

of HCl and then remember we’re still going to use our concentrated HCl

molarity that we found in Part A which was that 11.7 molarity of HCl. Now we’re

asked to find grams of a completely different compound and in this

case its sodium bicarbonate. So we’re gonna do a little bit of stoichiometry

here. So before I move on to stoichiometry, one thing to note is we

need our balanced equation. Because we’re going from moles or really one compound

of HCl to a totally different compound of sodium bicarbonate, we have to do a

mole to mole ratio and where do we get a mole to mole ratio, our balanced equation,

those coefficients in front. So our balanced equation would be HCl, plus

sodium bicarbonate and what’s gonna happen is that sodium or NA would then

combine with Cl so we’ll see that here these two would combine to form that

NaCl and then the next portion here is a little bit different because we’re gonna

form two additional products and this is from that hydrogen combining with that

bicarbonate portion. So when hydrogen gets combined with bicarbonate it forms

water and carbon dioxide. Okay now if we were to add everything up we’ll see that

this is already balanced, how nice. So everything has a 1, every single

compound has a 1 in front. So it’s all a one-to-one relationship. Now let’s start

this problem off with our given. So our given was that 1.75 liters of HCl.

Remember molarity is usually used as a conversion factor which is exactly what

we’re gonna use it for with this question. Now we’re gonna have 1.75 liters of HCl,

align our units so they can cancel. So 1 liter of HCl and where I’m taking

this by the way, I can rewrite this as 11.7 moles of HCl divided

by 1 liter of HCl. So that’s what you want to do whenever you’re looking at

molarity, split it apart into moles divided by liters. Now we’re at moles of

HCl on top and then I want those moles of HCl to cancel to then get to moles

of a completely different compound or moles really of our sodium bicarbonate.

So this is where our balanced equation will come in. We’re gonna do a mole to mole

ratio again and then remember everything had a 1

in front of the compound so I’m just gonna do 1 mole of HCl on top would be

1 mole of sodium bicarbonate. So moles of HCl would then cancel and then

finally we’re asked to find the grams of sodium bicarbonate, so we need to use the

molar mass of sodium bicarbonate. That’s what we’re gonna calculate right now. So

sodium has a mass of 22.99. Hydrogen has a mass of 1.01. Carbon 12.01

and oxygen is 16 however there are 3 oxygen within this compound so I’m gonna

multiply that by 3 and we’ll get 48. Now we’ll add everything together,

so adding all of these masses that we found together we will get the molar

mass of 84.01 grams per mole which we have to place on top. Since

we’re looking for grams, that’s why replacing grams on top and then remember

we want our units to cancel so moles of sodium bicarbonate, let’s align those

across from each other so they can cancel and finally we have our grams of

sodium bicarbonate. Multiply straight across since really we’re dividing by

1 and then from here our answer would be 1,720 grams of sodium

bicarbonate. And that was our pact example of a solution stoichiometry

question. If this video helped you out make sure to LIKE and subscribe and for

more resources that can help you with either molarity or stoichiometry or

really chemistry in general make sure to check out the description box and keep

practicing guys, keep going. You are absolutely capable. I’ll see you in the

next video.