1. A mixture of oxygen and neon gases, in a 7.19 L flask at 29 °C, contains 5.31 grams of oxygen and 1.34 grams of neon. The partial pressure of neon in the flask is atm and the total pressure in the flask is ______ atm.

2. A mixture of hydrogen and xenon gases is maintained in a 8.04 L flask at a pressure of 2.71 atm and a temperature of 11 °C. If the gas mixture contains 0.635 grams of hydrogen, the number of grams of xenon in the mixture is ____ g.

3. The stopcock connecting a 2.82 L bulb containing krypton gas at a pressure of 5.38 atm, and a 9.43 L bulb containing methane gas at a pressure of 1.31 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is _____ atm.

5. A mixture of oxygen and neon gases, at a total pressure of 608 mm Hg, contains 5.30 grams of oxygen and 1.34 grams of neon. What is the partial pressure of each gas in the mixture?

Po2 = _____ mm Hg

PNe = _____ mm Hg

6. A mixture of **hydrogen** and **xenon** gases contains **hydrogen**at a partial pressure of **279** mm Hg and **xenon** at a partial pressure of **549** mm Hg. What is the mole fraction of each gas in the mixture?

X_{H}** _{2}** = _____

X

**= _____**

_{Xe}7. A mixture of krypton and methane gases at a total pressure of 768 mm Hg contains krypton at a partial pressure of 190 mm Hg. If the gas mixture contains 5.20 grams of krypton, how many grams of methane are present?

Mass = g CH4

8. What volume of hydrogen gas is produced when 66.7 g of sodium reacts completely according to the following reaction at 25 °C and 1 atm?

sodium + water ——> sodium hydroxide + hydrogen

Volume = _____ L

9. How many grams of **iron **are needed to completely consume**60.6 **L of **chlorine gas **according to the following reaction at 25 °C and 1 atm?

**iron **(** s **) **+ chlorine ( g ) iron(III) chloride **(** s **)

______ grams **iron**

10. What is the total volume of gaseous products formed when 132 liters of bromine trifluoride react completely according to the following reaction? (All gases are at the same temperature and pressure.)

bromine trifluoride —–> bromine + fluorine

Volume = _____ L

11. What volume of **nitrogen monoxide** is produced when **184**liters of **nitrogen gas** react according to the following reaction? (All gases are at the same temperature and pressure.)

**nitrogen**(**g**) **+ oxygen(g)** **nitrogen monoxide**(**g**)

__________ liters **nitrogen monoxide**

### Step 1 of 2

Given data:

- Volume (
*V*) = 19L - Temperature (
*T*) = 29°C - Mass of oxygen (�O2) = 31g
- Mass of neon (�Ne) = 34g

Find: The partial pressure of neon in the flask is atm and the total pressure in the flask is ______ atm.

**Explanation:**

To solve this problem, we need to use the ideal gas law equation:

PV = nRT

### Step 2 of 2

To solve this problem, we’ll follow these steps:

- Convert the masses of oxygen and neon to moles.
- Use the ideal gas law to find the moles of each gas.
- Calculate the partial pressure of neon using its moles and the total pressure.
- Find the total pressure in the flask.

**Step 1: Convert masses to moles.**

Moles of O2=mO2Molar mass of O2

Moles of Ne=mNeMolar mass of Ne

The molar masses are:

- Molar mass of O2=32 gmol−1
- Molar mass of Ne=20.18 gmol−1

**Step 2: Use the ideal gas law to find moles.**

PV=�RT

n=RTPV

**Step 3: Calculate the partial pressure of neon.**

Partial pressure of Ne=nNeRTV

**Step 4: Find the total pressure.**

Total pressure = Partial pressure of O2+Partial pressure of Ne

Let’s calculate these values:

**Step 1:**

Moles of O2=325.31

Moles of Ne=20.181.34

**Step 2:**

For O2

PO2=nRTV

PO2=5.32×.8,206×30232×7.19PO2=0.5719atm

For Ne

PNe=nRTV

PNe=1.34×0.08206×30220.18×7.19PNe=0.2288atm

**Step 3:**

Total pressure=Partial pressure of O2+Partial pressure of Ne

Pt=PO2 +PNe=0.5719 +0.2288=0.8007 atm

**Explanation:**

Hence, by using Ideal gas equation we get P(t)= 0.8007 atm.

##### Final solution

Therefore, by adding both P(O2) and P(Ne) we got P(t) as 0.8007 atm.