10.4: Gas Mixtures
- Page ID
- 170548
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Exercise \(\PageIndex{1}\)
What is the volume occupied by a mixture of 0.595 mol of N2 gas and 0.685 mol of O2 gases at 1.75 atm and 23.2°C?
- Answer
-
17.9 L
\(\lV=\frac{nRT}{P}\)
n = 0.595 + 0.685 = 1.28 moleT = 23.2°C + 273.15 = 296.35 K
P = 1.75 atm; R = 0.08205 L atm mol-1 K-1
\(\V = \frac{1.28*0.08205*296.35}{1.75} = 17.88\;L\)
Exercise \(\PageIndex{2}\)
A vessel with a volume of 20.5 L contains 79.9 g of nitrogen gas, 2.80 g of hydrogen gas, and 0.403 g of argon gas. At 25°C, what is the pressure in the vessel?
- Answer
-
5.07 atm
Exercise \(\PageIndex{3}\)
Water can be decomposed by electrolysis into hydrogen gas and oxygen gas. What mass of water must decompose to fill a 3.00 L flask to a total pressure of 2.00 atm at 298 K with a mixture hydrogen and oxygen? (R = 0.08206 L⋅atm/mol⋅K)
2 H2O(l) → 2 H2(g) + O2(g)
- Answer
-
2.95 g
Exercise \(\PageIndex{4}\)
Carbon monoxide reacts with oxygen to form carbon dioxide.
2 CO(g) + O2(g) → 2 CO2(g)
In a 1.00 L flask, 4.30 atm of CO reacts with 2.50 atm of O2. Assuming that the temperature remains constant, what is the final pressure in the flask?
- Answer
-
4.95 atm
Mole Fraction and Partial Pressure
Exercise \(\PageIndex{5}\)
The partial pressures of CH4, N2, and O2 in a sample of a gas mixture were found to be 191 mm Hg, 451 mm Hg, and 583 mm Hg, respectively. Calculate the mole fraction of each gas.
- Answer
-
\(\large \chi _{CH_{4}} = 0.156\;,\;\chi _{N_{2}} = 0.368\;,\;\chi _{O_{2}} = 0.476\)
Total Pressure = 191 + 451 + 583 = 1225 mm Hg
\(\large \chi _{CH_{4}} = \frac{\rho _{CH_{4}}}{P}=\frac{191}{1225} = 0.1559\)
\(\large \chi _{N_{2}} = \frac{\rho _{N_{2}}}{P}=\frac{451}{1225} = 0.3681\)
\(\large \chi _{O_{2}} = \frac{\rho _{O_{2}}}{P}=\frac{583}{1225} = 0.4759\)
Exercise \(\PageIndex{6}\)
A gaseous mixture containing 7.60 mol Ar and 7.40 mol CO2 has a total pressure of 3.50 atm. What is the partial pressure of CO2?
- Answer
-
\(\large \rho _{CO_{2} = 1.73 atm\)
\(\large \chi _{Ar} = \frac{7.60}{7.60+7.40}=0.507\)
\(\large \chi _{CO_{2}} = \frac{7.40}{7.60+7.40}=0.493\)
\(\large \rho _{CO_{2}= P_{total}\;*\;\chi _{CO_{2}}=3.50\;*\;0.493=1.726\; atm\]
Exercise \(\PageIndex{7}\)
A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715, what is the mass percent of Ar?
- Answer
-
86.2%