Now here we're going to talk about gas stoichiometry. But first, recall that stoichiometry deals with the numerical relationship between compounds in a balanced chemical equation. And when we say gas stoichiometry, that deals with stoichiometric calculations of chemical reactions that produce gases. Now, when talking about gas stoichiometry, we have to employ our stoichiometric chart. Within this chart, we are going to use the given quantity of a compound, in this case, possibly a gas, to determine the unknown quantity of another compound within our balanced chemical equation. Now that we know what gas stoichiometry is and how it connects to an ideal stoichiometry which we've seen before, click on to the next video, and let's take a better look at this stoichiometric chart.
Gas Stoichiometry - Online Tutor, Practice Problems & Exam Prep
Gas Stoichiometry involves chemical reactions that contain gases.
Gas Stoichiometry
Gas Stoichiometry
Video transcript
Gas Stoichiometry
Video transcript
So here we're going to take a look at this stoichiometric chart. Here we have 4 moles of silver solid reacting with 1 mole of oxygen gas to produce 2 moles of silver oxide solid. They give us the volume, pressure, and temperature of our oxygen gas, and we're asked to determine the grams of our silver oxide. Realize here that if we're thinking of the ideal gas law, PV=nRT , by giving us the pressure, volume, and temperature, we can isolate the moles of that particular gas. So, remember n= P⋅V R⋅T . So we would say that giving us pressure volume over RT, since it's given to us at the amount given, that would directly feed into moles of given. Or they could give us the grams of one of the other compounds or elements within a balanced reaction and so we go from grams of given to moles of given. Going from moles of given to moles of unknown requires us kind to have a leap of faith. Going from an area where we know some information, our given information, to an area where we know nothing at all, our unknown information. Because of this leap of faith, we call this the jump. As we go from our given region to our unknown region. Now remember with stoichiometry, when we go and make this jump, we have to do a mole-to-mole comparison, so we use the coefficients in the balanced equation. From this point, if we know the moles of our unknown we can easily transform it into ions, atoms, formula units, molecules, or even back into grams, but now for the unknown. If you're not quite familiar with this stoichiometric chart, make sure you go back and take a look at my videos on stoichiometry. This is where we first laid down the groundwork for our stoichiometric chart and this is just a slight modification to that previous one. Now that we've seen the stoichiometric chart, we'll put it into action as we start doing questions dealing with gas stoichiometry.
Gas Stoichiometry Example 1
Video transcript
So here it says, what mass of silver oxide is produced when 384 milliliters of oxygen gas at 736 millimeters of mercury and 25 degrees Celsius is reacted with excess solid silver. Alright. So step 1 is we need to map out the portion of the stoichiometric chart you will use. They are giving us within this question the volume of the gas, the pressure of the gas, and its temperature. So we know with that information we can find the moles of our gas because moles equal PVRT. First, we need to convert our pressure into atmospheres. Remember, 1 atmosphere is 760 torrs or 760 millimeters of mercury. Millimeters of mercury cancel out, and when we do that, we get 0.9684 atmospheres. Then we are going to take the volume, which is 0.384 liters. You're going to divide that by the R constant. And then remember you add 273.15 to get 298.15 Kelvin from the temperature. This information here, what is it doing? It's converting the given quantity that we have, these amounts, into our moles of given. So when I plug all of this in, my moles given for oxygen gas comes out to be 0.015199 molesO2gas. Now that we have moles given, we go to step 3. We do a mole to mole comparison to convert moles of given into the moles of our unknown. So we're going to say here, moles of oxygen go on the bottom. Our unknown is what we're being asked to find, which is our silver oxide. At this point, we need to do a mole to mole comparison, which says that for every one mole of oxygen gas we have 2 moles of silver oxide. So moles of oxygen gas cancel out, and now we have moles of silver oxide. Now, step 4, if necessary, convert the moles of unknown into the final desired units. Alright. They're not asking us for moles of silver oxide, so I'm just continuing onward till we get grams of silver oxide. So, for every one mole of silver oxide, the mass of 2 silvers and one oxygen has a combined mass of 231.7 grams of silver oxide. Moles of silver oxide cancel out, and now I have my final answer of silver oxide. So that comes out to be 7.0 grams of silver oxide. Here our answer has 2 significant figures because 25 has 2 significant figures. Here we didn't have to do step 5 because in step 5, it says recall, if you calculate more than one final amount then you must compare them to determine the theoretical yield. Here we are only given amounts for oxygen gas. So step 5 isn't necessary. Now if you don't remember the whole concept of theoretical yield, make sure you go back and take a look at my topic videos on theoretical yield. What does it mean and how does it relate to stoichiometry is explained in those series of videos. Alright. So now that we've gotten our answer, this is the approach we need to take when it comes to gas stoichiometry.
The metabolic breakdown of glucose (C6H12O6) (MW:180.156 g/mol) is given by the following equation:
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Calculate the volume (in mL) of CO2 produced at 34°C and 1728.9 torr when 231.88 g glucose is used up in the reaction.
8.6×105 mL
6.8×104 mL
8.6×104 mL
4.3×104 mL
The oxidation of phosphorus can be represented by the following equation:
P4 (s) + 5 O2 (g) → 2 P2O5 (g)
If 1.85 L of diphosphorus pentoxide form at a temperature of 50.0 ºC and 1.12 atm, what is the mass (in g) of phosphorus that reacted?
Determine the mass (in grams) of water formed when 15.3 L NH3 (at 298 K and 1.50 atm) is reacted with 21.7 L of O2 (at 323 K and 1.1 atm).
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
12 g
19 g
25 g
38 g
51 g
