Now here we're gonna 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're 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 ideal stoichiometry which we've seen before, click on to the next video, and let's take a better look at this stoichiometric chart.

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# Gas Stoichiometry - Online Tutor, Practice Problems & Exam Prep

Gas stoichiometry involves calculating the relationships between gases in balanced chemical equations. Using the ideal gas law, PV=nRT, we can determine moles of gases from given pressure, volume, and temperature. Transitioning from known to unknown quantities requires a mole-to-mole comparison based on coefficients in the balanced equation. This process is essential for converting moles of reactants to products, facilitating accurate predictions of yields in gas-evolution reactions.

**Gas Stoichiometry** involves chemical reactions that contain gases.

## Gas Stoichiometry

### Gas Stoichiometry

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### Gas Stoichiometry

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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. Here 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 here remember moles equals pressure times volume divided by RT. So we would say that giving us pressure volume over RT, since it's given to us at the amount of given, that would directly feed into moles of given. Or they could give us the grams of 1 of the other compounds or elements within a balanced reaction and so we go from grams of given still to moles of given. Going from moles of given to moles of unknown requires us to 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

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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'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 RT. First, we need to convert our pressure into atmospheres. Remember, 1 atmosphere is 760 torres 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 this in, my moles of given for oxygen gas comes out to be 0.015 199 moles of oxygen gas. Now that we have moles of 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 gonna 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 (C_{6}H_{12}O_{6}) (MW:180.156 g/mol) is given by the following equation:

C_{6}H_{12}O_{6} (s) + 6 O_{2} (g) → 6 CO_{2} (g) + 6 H_{2}O (l)

Calculate the volume (in mL) of CO_{2} produced at 34°C and 1728.9 torr when 231.88 g glucose is used up in the reaction.

#### Problem Transcript

The oxidation of phosphorus can be represented by the following equation:

P_{4} (s) + 5 O_{2} (g) → 2 P_{2}O_{5} (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 NH_{3 }(at 298 K and 1.50 atm) is reacted with 21.7 L of O_{2} (at 323 K and 1.1 atm).

4 NH_{3} (g) + 5 O_{2} (g) → 4 NO (g) + 6 H_{2}O (g)

### Here’s what students ask on this topic:

What is gas stoichiometry and how is it used in chemical reactions?

Gas stoichiometry involves calculating the relationships between gases in balanced chemical equations. It uses the ideal gas law, $PV=nRT$, to determine the moles of gases from given pressure, volume, and temperature. By knowing the moles of a gas, we can use stoichiometric coefficients from the balanced equation to find the moles of other reactants or products. This process is essential for predicting yields in reactions that produce or consume gases.

How do you use the ideal gas law in gas stoichiometry calculations?

In gas stoichiometry, the ideal gas law $PV=nRT$ is used to find the number of moles of a gas. Given the pressure (P), volume (V), and temperature (T) of a gas, you can rearrange the equation to solve for moles (n): $n=\frac{PV}{RT}$. Once you have the moles, you can use stoichiometric coefficients from the balanced chemical equation to find the moles of other substances involved in the reaction.

What is the stoichiometric chart and how is it used in gas stoichiometry?

The stoichiometric chart is a tool used to visualize and organize the steps in stoichiometric calculations. It helps in converting given quantities (like grams or volume) to moles, and then using mole-to-mole ratios from the balanced equation to find unknown quantities. In gas stoichiometry, the chart incorporates the ideal gas law to convert between volume, pressure, temperature, and moles of gases. This systematic approach ensures accurate calculations and helps in understanding the relationships between reactants and products in a chemical reaction.

How do you convert between moles of a gas and grams of a solid in a chemical reaction?

To convert between moles of a gas and grams of a solid in a chemical reaction, follow these steps: 1) Use the ideal gas law $PV=nRT$ to find the moles of the gas. 2) Use the stoichiometric coefficients from the balanced chemical equation to convert moles of the gas to moles of the solid. 3) Finally, use the molar mass of the solid to convert moles to grams. This process ensures accurate conversion between different states of matter in a reaction.

What role do stoichiometric coefficients play in gas stoichiometry?

Stoichiometric coefficients in a balanced chemical equation represent the mole ratios of reactants and products. In gas stoichiometry, these coefficients are crucial for converting between moles of different substances. For example, if the equation shows that 1 mole of gas A reacts with 2 moles of gas B to produce 1 mole of gas C, the coefficients (1:2:1) guide the mole-to-mole conversions. This allows for accurate predictions of the amounts of reactants needed or products formed in a gas-evolution reaction.