Gas Laws Advanced Problem Solving. Visit: //checkin.ics.uci.edu/ Log in and select Chem 1A. When...

Gas Laws

Advanced Problem Solving

• Visit: https://checkin.ics.uci.edu/

• Log in and select Chem 1A.

• When prompted, type the word of the day: tougher• Ensure when asked if you will share your location you select “allow”.

• Visit: https://learningcatalytics.com/

• Sign in MasteringChemistry account name

• When prompted, type session ID:28491493

Please turn off all downloads. You can usually pause them by clicking on them and clicking pause. Facebook and youtube are officially banned during class time

(this is not me micromanaging, it’s a legit internet issue as told to me by OIT).

Remember back 5 rows of even side

are no seating zones

Visit: https://checkin.ics.uci.edu/ Word of the day: tougherLearningCatalytics: session ID:28491493

https://checkin.ics.uci.edu/


https://learningcatalytics.com/


SURVEY QUESTIONS• How to know what specific formulas to use for specific gas law problems

• Use PV=nRT if only one set of conditions are given, in which case nothing cancels.

• Use ratio form, if two sets of conditions are given. Cancel anything that is constant.

• Will we has problems that use limiting reagents and ideal gas laws? Will the final contain ideal gas law problesm?

• Absolutely and yes. The final will contain everything.

• A myriad of questions about conversions between moles and grams, mol ratios, and using chemical reactions for mol ratios.

• Please come see me/ Paul/ Tutors in office hours.

• Do we need to know {insert whatever conversion you are wondering about here}?

• No, they are on the front page of the exam equation sheet that you already have.

𝑃1𝑉 1𝑛1𝑇 1

=𝑃 2𝑉 2𝑛2𝑇 2



DALTON’S LAW• Dinitrogen oxide gas was generated from the thermal

decomposition of ammonium nitrate and collected over water. The wet gas occupied 126 mL at 21 C when the atmospheric pressure was 755 Torr. What volume would the same amount of dry Dinitrogen oxide have occupied if collected at 755 Torr and 21 C. The vapor pressure of water is 18.65 Torr at 21 C.

Hint 1: If the total pressure is 755 Torr, what is the pressure of N2O?

Hint 2: How do you determine the volume of a gas at a new pressure?

Pressure of N2O:

Atmospheric =P(N2O)+P(H20vapor)

755-18.65 torr=736.35 torr

Then use:

V2=123mL



REACTIONS: SINGLE GAS PRODUCT• Suppose that 200.0 mL of propane gas C3H8 at 1.00 atm and 298K is mixed

with 1.00 L of oxygen gas at the same pressure and temperature and burned to form carbon dioxide gas and liquid water. Determine the final volume of the reaction mixture at 2.00 atm and 298 K if the reaction goes to completion. (note: volume of liquid water is negligible).

Hint1: Write out the equation and balance it.

Balance Carbons and Hydrogens first then oxygens. 1C3H8:5O2

mol ∝ V mol ratio= volume ratio

(At constant temp/pressure)Visit: https://checkin.ics.uci.edu/ Word of the day: tougherLearningCatalytics: session ID:28491493


You may fill in and solveORPressure doubles, so volume…….HalvesVolume=300.mL

REACTIONS: • Suppose that 200.0 mL of propane gas C3H8 at 1.00 atm and 298K is mixed

with 1.00 L of oxygen gas at the same pressure and temperature and burned to form carbon dioxide gas and liquid water. Determine the final volume of the reaction mixture at 2.00 atm and 298 K if the reaction goes to completion. (note: volume of liquid water is negligible).Hint 2: Check for any

limiting reagents, which may leave some gaseous reactant unreacted. 1C3H8 : 5O2

mol ∝ V mol ratio= volume ratio

(At constant temp/pressure)

200.0mL of propane requires 1.00L of O2

We have exactly that, no excess reagent,

reaction goes to C3H8=O2=0atm

You could solve for moles of CO2 here, but that would be a lot of calculations and a lot of time/possibilities for mistakes. Instead look for how much volume would be made at the SAME temp and pressure, and then alter it for the new temperature and pressure.

Short way: more thinking, more logic jumps, less math



REACTIONS: SINGLE GAS PRODUCT• Suppose that 200.0 mL of propane gas C3H8 at 1.00 atm and 298K is mixed with 1.00 L of oxygen gas at the same

pressure and temperature and burned to form carbon dioxide gas and liquid water. Determine the final volume of the reaction mixture at 2.00 atm and 298 K if the reaction goes to completion. (note: volume of liquid water is negligible).

Long way: less thinking, less logic jumps, more math

𝑛𝐶 3𝐻 8=𝑃𝑉𝑅𝑇

=1.00𝑎𝑡𝑚∗0.2000𝐿

0.0821𝐿∗𝑎𝑡𝑚𝑚𝑜𝑙∗𝐾

∗298𝐾=0.008175𝑚𝑜𝑙𝐶3𝐻 8

𝑛𝑂 2=𝑃𝑉𝑅𝑇

=1.00𝑎𝑡𝑚∗1.00𝐿

0.0821𝐿∗𝑎𝑡𝑚𝑚𝑜𝑙∗𝐾

∗298𝐾=0.040873𝑚𝑜𝑙𝑂 2

∗3𝑚𝑜𝑙𝐶𝑂21𝑚𝑜𝑙𝐶3𝐻 8

=0.245𝑚𝑜𝑙

∗3𝑚𝑜𝑙𝐶𝑂25𝑚𝑜𝑙𝐶 3𝐻 8

=0.245𝑚𝑜𝑙

Same mol created. Both are used up! 0.245mol CO2 created!

𝑉 𝐶𝑂 2=𝑛𝑅𝑇𝑃

=0.245𝑚𝑜𝑙∗0.0821

𝐿∗𝑎𝑡𝑚𝑚𝑜𝑙∗𝐾

∗298𝐾

2.00𝑎𝑡𝑚=0.300𝐿𝐶𝑂 2



REACTIONS: • The reaction of solid dimethylhydrazine, (CH3)2N2H2, and liquefied

dinitrogen tetraoxide, has been investigated as a rocket fuel; the reaction produces gaseous carbon dioxide, nitrogen and water vapor, which are ejected as exhaust gases. In a controlled experiment solid dimethylhydrazine reacted with excess nitrogen tetroxide with a final pressure of 2.50 atm, with a temperature of 400.0K. Find the pressure of each gas. Type the total pressure into learning catalytics.

Hint: Write out and balance the equation.

Hint2: what do you think must be true of the ratio of the gases pressures? Remember n∝PThey must equal the mole ratios coefficients: 4 H2O: 2 CO2: 3N2

Hint3: What is the total pressure? What are the mol ratios of products?

H2O=(4/9)*(2.50 atm)=1.11atm CO2=(2/9)*(2.50 atm)=0.556 atm

N2=(3/9)*(2.50atm)=0.834 atm

H2O

CO2

N2



• A 1.00 L sample of chlorine gas at 1.00 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the same temperature and pressure. There is a single gaseous product, which fills a 2.00L flask at 1.00 atm and 298 K. Use this information to determine the following characteristics of the product. A) its empirical formula, B) its molecular formula, C) the most favorable Lewis structure based on formal charge arguments (the central atom is an N atom); d) the molecular shape.

Hint 1: (a)“reacts completely” in this question means no limiting reagent, everything reacts all the way to zero.

Hint 2: (a) Remember V ∝ n at constant pressure. Also don’t forget the reactants are diatomic gases.

Hint 3: (b) You know P, V, T of the final reaction flask, what is n?, what mass of reactants were put in? Where did all those reactants go. What is the units of molecular mass?

Ratio of gas moles=ratio of gas volume: 1Cl2:1N2:2O2

ThenTurn into g using molar mass

To find total grams: do the following for each reactant:

Plan: find total grams, find total moles. Do grams/moles.

To find final total moles: use ideal gas law for final reaction flask.

A)

B)



• A 1.00 L sample of chlorine gas at 1.00 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the same temperature and pressure. There is a single gaseous product, which fills a 2.00L flask at 1.00 atm and 298 K. Use this information to determine the following characteristics of the product. A) its emperical formula, B) its molecular formula, C) the most favorable Lewis structure based on formal charge arguments (the central atom is an N atom); d) the molecular shape.

Find total gPut in by doing each reactant separately:

1.15g+2.90g+2.62g=6.67g totalVisit: https://checkin.ics.uci.edu/ Word of the day: tougherLearningCatalytics: session ID:28491493


MORE REACTIONS:(I’m not sure we’ll make it to this

one, if we don’t the answer will be posted online)


Ratio of gas moles=ratio of gas volume: 1Cl2:1N2:2O2

Total Mass (from other slide) =6.67g total

Total Moles reactant:

Compare to mass of NClO2=14+35+32= 81g/mol= same, so NClO2 is also the molecular

formula




Why can’t we add another double bond to minimize formal charge further?

Steric number 3: trigonal planar.



Gas Laws Advanced Problem Solving. Visit: //checkin.ics.uci.edu/ Log in and select Chem 1A. When...

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