Chem L3 - Partial pressure=P initial*X 02 Χ=mole fraction= mols O2total mols In a flask a student inserts 400mL of CO 2(g 300 K and 1 atm then

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Chemistry Lecture #3- Midterm #2 February 2, 2011 Average velocity = 3RTMM , where MM= molar mass and R= 8.021 Steeper slope with Low temperature, high MM, high density Smoother slope with High temperature, low MM, low density A student combusts 114.2g of octane, C 8 H 18 (l) at 1.00 atm and 298 K. How many liters of CO 2 gas is produced? C 8 H 18 (l)+ 252 O 2 (g) 8CO 2 (g) + 9H 2 O(g) 114.2g=1mol octane=8mol CO 2 then use PV=nRT to find liters V= nRTP V=(8*0.0821*298)/1= 195.49L of CO 2 Partial Pressures: What is the pressure in the flask? What is the pressure due to the O 2 gas?
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Unformatted text preview: Partial pressure=P initial *X 02 Χ=mole fraction= mols O2total mols In a flask a student inserts 400mL of CO 2 (g) @ 300 K and 1 atm, then inserts 571mL of Xe @ 327 K and 1atm. What is the pressure inside the 2.0 L flask at 273 K? P total = Pi , where P i is the Partial Pressure Step 1: CO 2 n= PVRT =( * . . * 1 atm 0 400L0 0821 300 ) = 0.0162 moles Step 2: Xe n=( * . . * 1atm 0 571L0 0821 327 ) = 0.0213 moles Step 3: Calculate p inside flask P= nRTV P= . + . . ( ) 0 0162 0 02130 0821 273 2L P=0.42 atm...
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This note was uploaded on 02/04/2011 for the course CH 222 taught by Professor Haak during the Spring '09 term at Oregon State.

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