Start typing, then use the up and down arrows to select an option from the list. ## General Chemistry

Learn the toughest concepts covered in Chemistry with step-by-step video tutorials and practice problems by world-class tutors

7. Gases

# Gas Stoichiometry

Gas Stoichiometry involves chemical reactions that contain gases.

Gas Stoichiometry
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concept

## Gas Stoichiometry 47s
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Now here, we're gonna talk about gasto geometry. But first recall that stored geometry deals with the numerical relationship between compounds in a balanced chemical equation. And when we say gas Tokuyama tree that deals with Stoke geometric calculations of chemical reactions that produce gas is now when talking about Gastaut geometry, we have to employ our story geometric chart. Within this chart, we're gonna 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 gastric geometry is and how it connects to an idea of stork geometry which we've seen before, click on to the next video and let's take a better look at this stock your metric chart.
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concept

## Gas Stoichiometry 2m
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So here we're gonna take a look at this toy Geometric chart. Here we have four moles of silver solid reacting with one mole of oxygen gas to produce two moles of silver oxide solid. Here they give us the volume, pressure and temperature off our oxygen gas and were asked to determine the grams of our silver oxide. Realize here that if we're thinking of the ideal gas law, PV equals NRT By giving us the pressure, volume and temperature, we can isolate the moles of that particular gas. So here remember, moles equals pressure times volume divided by R t. So we would say that giving us pressure volume over rt since it's given to us that the amount of 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 have given still two moles of given going from moles of Given two moles of unknown requires us kinda have a leap of faith going from an area where we know some information, are given information to an area where we know nothing at all are unknown information. Because of this leap of faith, we call this the jump as we go from are given region toe are unknown region. Now remember what stoke geometry. When we go on, 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 could easily transform it into ions, Adams, formally units, molecules or even back into grams. But now for the unknown. If you're not quite familiar with the story Geometric chart, make sure you go back and take a look at my videos on Sony geometry. This is where we first laid down the groundwork for our stock geometric chart. And this is just a slight modification to that previous one. Now that we've seen this documentary chart will put into action as we start doing questions dealing with gas stock. Yama
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example

## Gas Stoichiometry Example 1 3m
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So here it says, What mass of silver oxide is produced when 384 millimeters of oxygen gas at 736 millimeters of mercury and 25 degrees Celsius is reacted with excess solid silver. Alright, so step one is we need to map out the portion of the story geometric chart you will use. They're 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 equals pressure. Times volume divided by R T for us. We need to convert our pressure into atmospheres. Remember, one atmosphere is 760 tours for 760 millimeters of mercury millimeters of mercury cancel out. And when we do that, we get 0.9684 atmospheres. Then we're going to take the volume which is 0.384 leaders. You're going divide that by the are constant and then remember, you add to 73.15 to get to 98.15 Kelvin from the temperature. This information here what is it doing? Its converting the given quantity that we have These amounts into our moles of given So when I plug all this in my moles of given for oxygen gas comes out to the 0.15199 Moles of oxygen gas Now that we have moles of given we go to step three we do a multi mode comparison to convert moles have given into the moles of are unknown So we're gonna say here moles of oxygen go on the bottom are 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 two moles of silver oxide So moles of oxygen gas cancel up and now we have moles of silver oxide. Now step four 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 So where we get grams of silver oxide? So for every one mole of silver oxide, the mass of two silvers and one oxygen has a combined mass off to 31 0.7 g of silver oxide moles of silver oxide. Cancel out, and now I'll have my final answer of silver oxide. So that comes out to be 7.0 grams off silver oxide. Here are answer has two significant figures because 25 has two significant figures here. We didn't have to do Step five, because from Step five, it says Recall. If you calculate more than one final amount thing, you must compare them to determine that theoretical yield. Here. We're only given amounts for oxygen gas, so Step five isn't necessary. Now, if you don't remember the whole concept of theoretical healed, make sure go back and take a look at my topic. Videos on Theoretical Yield What does it mean? And how does it relate to stoke? Geometry is explained in those series of videos. All right, so now that we've gotten our answer, this is the approach we need to take when it comes to gas store geometry.
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Problem

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.

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Problem

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?

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Problem

Determine the mass (in grams) of water formed when 15.3 L NH(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) 