Hello hello Melissa Maribel here with some examples on stoichiometry. Before we begin, remember there are three different types of conversion factors for stoichiometry. The first type is molar mass. We use molar mass whenever we're converting from grams to moles or moles to grams. The second one is a mole ratio which is found on your balanced equation. You use a mole to mole ratio whenever you are changing your compound. And the last one is Avogadro's number. You use Avogadro's number whenever you see the keywords of atoms, molecules, particles, and formula units. All right, now that we know that, let's jump right in. What mass of hydrogen peroxide must decompose to produce 48.64 grams of water? And we're given our balanced equation. Hydrogen peroxide is that H2O2. And let's figure out what we're given as a whole. So what we're given is the 48.64 grams of water and we're asked to find our mass or our grams of hydrogen peroxide. And as I mentioned hydrogen peroxide is H2O2. So we will need this balanced equation because we're trying to get from grams of H2O to a completely different compound of grams of H2O2. Our plan is to go from grams of water to grams of H2O2. Now we can't just instantly go to grams to grams. We have to go from grams to moles and then moles to grams. So our first step is to convert our grams of water to moles of water using our molar mass. Whenever we go from grams to moles we use the molar mass of that compound. Then once we have our moles of H2O we will change our compound entirely using our mole to mole ratio found on our balanced equation. So we'll go from our moles of water to moles of hydrogen peroxide. Now that we have our moles of hydrogen peroxide we can then convert that down to our grams of hydrogen peroxide. Once again using our molar mass of that compound. So let's set this up. You want to put your given on top, that's what you always start with is your given. So we have our 48.68 grams of water and we're going to use that first conversion factor as our molar mass of water and it's going to be 18.02 grams. You align these two across from each other because then we will then cancel out our grams of water. We now are left with moles of water and we're going to go back to our balanced equation to do a mole to mole ratio. So looking back at our balanced equation we see that we have two moles of hydrogen peroxide, so that's why I placed a 2 here, and for every two moles of water. That's also why put the 2 moles on the bottom so they can then cancel those moles of water would then cancel and we're at moles of hydrogen peroxide. Now from moles of hydrogen peroxide we will then use our last conversion factor, the molar mass of hydrogen peroxide to get two grams of hydrogen peroxide because our moles would then cancel and we would be left with just the grams of hydrogen peroxide which is what we're solving for. So multiply straight across, divided by that 18.02 times the 2 and you get 91.83 grams of hydrogen peroxide. Our next example states, how many molecules of carbon dioxide are required to react with a hundred and seventy four grams of carbon monoxide? So remember that carbon dioxide, that "di" meaning 2 is our CO2. carbon monoxide, "mono" meaning 1, is our CO. So let's identify what we're given and what we're solving for. So we're given that 174 grams of carbon monoxide or CO and we are finding molecules of CO2, key word right there, molecules. So remember anytime you see that word molecules you use Avogadro's number. Also, we're changing our compound so we have to use a mole to mole ratio using this balanced equation. Let's set it up. Setting up our plan we see that we're going from grams of CO or carbon monoxide and we're trying to get to molecules of carbon dioxide. So we're changing our compound, we have to convert first to moles of carbon monoxide. So from going from grams to moles we use the molar mass of carbon monoxide, that's our first conversion factor. Then, now that we're at moles of carbon monoxide, we will change our compound to the moles of carbon dioxide using our balanced equation using that mole to mole ratio. Now that we're at moles of carbon dioxide we can then convert this to molecules of carbon dioxide using Avogadro's number. Your proper set up would then begin with our grams of carbon monoxide, because that was our given, and then aligning those grams across from each other so they can then cancel. We would then get moles of our carbon monoxide. Next we're right here on our second conversion factor where we have to use our balanced equation. So these numbers, if we go back to our balanced equation, we would then see that that 2 moles of carbon dioxide came right here from the coefficient from our balanced equation. This other 2 came from right here the other 2 on our balanced equation. That's where we get our mole to mole ratio from. Align those moles across from each other, specifically the moles of carbon monoxide, and those would cancel. Now we're at moles of CO2 and then we want to cancel out our moles of CO2 to get to molecules of CO2. So using Avogadro's number on top, we'll see that the moles of carbon dioxide would cancel and we would be left with molecules of CO2. Make sure to multiply straight across and divide by, these two multiplied, and you get 3.74 x 10^24 molecules of CO2. Now it's okay to need some more help and to need to see more practice problems or examples. That's actually exactly why I've done so many videos on this so check them out, I've gone into even more and more detail with more and more examples to really understand this topic because you will keep seeing stoichiometry in chemistry all throughout. Now if there's a specific type of question that you have on stoichiometry, feel free to leave a comment below and I'll answer it. 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