Using the Ideal Gas Law or any of the equations you found in questions A-‐D, answer the following questions about gases. Assume ideal behavior! Be sure to put all values into the SI units (pressure in atm, the temperature in K, volume in L).

Useful Conversion Factors and the Gas Constant:
Quantity

Pressure
1 atm = 760 torr
1 atm = 101 kPa
1 mmHg= 1 torr
Gas Constant R=0.0821 Latm/molK

1. A rigid container (25 mL) filled with oxygen gas has a pressure of 699 torr at a temperature of 25oC. How many moles of oxygen are in the container?

2. A 57.8-‐mL sample of gas in a cylinder is warmed from 20oC to 120oC. What is the volume at the final temperature?

Part II. Stoichiometry with Gaseous Reactants/Products
We now have a new way to find the moles of a compound! If a chemical is in the gaseous state, it is pretty hard (read: impossible) to measure the mass of the gas on a balance. Therefore, to determine the number of moles of a gaseous reactant (or product) in a chemical reaction, scientists measure the temperature, pressure, and volume of the gas. Using PV=nRT allows us to calculate the number of moles-‐ now ‘n’ can be used just like we always use it!

1. How many grams of Li3N can form when the following chemicals are allowed to react according to the reaction below? Excess solid Li reacts with 50 mL of nitrogen gas (1.23 atm, 293 K).

6 Li (s) + N2 (g) --> 2 Li3N (s)

1. A rigid container (25 mL) filled with oxygen gas has a pressure of 699 torr at a temperature of 25oC. How many moles of oxygen are in the container?

p*V = n*R*T

n = (p*V)/R*T

⇒with n = the number of moles oxygen = TO BE DETERMINED

⇒with p = the pressure of gas = 699 torr = 699/760 atm = 0.919737 atm

⇒with V = the volume of gas = 25 mL = 0.025 L

⇒with R = the gas constant = 0.08206 L*atm /mol*K

⇒with T = the temperature = 25 °C = 298 K

n = (0.919737*0,025)/(0.08206*298)

n = 0.000940 moles oxygen

A 57.8-‐mL sample of gas in a cylinder is warmed from 20oC to 120oC. What is the volume at the final temperature?

V1/T1 = V2/T2

⇒with V1 = the initial volume = 57.8 mL = 0.0578 L

⇒with T1 = the initial temperature = 20°C = 293 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the final temperature = 393 K

0.0578 L / 293 K = V2 / 393 K

V2 = (0.0578 / 293) * 393

V2 = 0.0775 L = 77.5 mL

The new volume is 77.5 mL

How many grams of Li3N can form when the following chemicals are allowed to react according to the reaction below? Excess solid Li reacts with 50 mL of nitrogen gas (1.23 atm, 293 K).

6 Li (s) + N2 (g) --> 2 Li3N (s)

p*V = n*R*T

n = (p*V)/(R*T)

⇒with n = the moles of nitrogen gas = TO BE DETERMINED

⇒with p= the pressure of nitrogen gas = 1.23 atm

⇒with V = the volume of nitrogen gas = 0.050 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 293 K

Moles nitrogen gas = (1.23 * 0.050) / (0.08206*293)

Moles nitrogen gas (N2) = 0.00256 moles

For 6 moles of Li we need 1 mol N2 to produce 2 moles Li3N

For 0.00256 moles N2 we'll have 2*0.00256 = 0.00512 moles Li3N

Mass Li3N = moles Li3N * molar mass Li3N

Mass Li3N = 0.00512 moles * 34.83 g/mol

Mass Li3N = 17.83 grams Li3N