Butane Burns as: 2C4H10(g) +13O2(g) --> 8CO2(g) + 10H2O(g)

Chemistry
Tutor: None Selected Time limit: 1 Day

How many liters of CO2(g) at (3.90 x 10^-4 atm) and (1.2 x 10^3 degrees C) from 5.60L of C4H10(g) at 1.50 atm and 25 degrees C?

May 19th, 2015

DATA:

Pressure for butane =1.50atm

Volume of butane= 5.60L

Temperature for butane= 27C0 = 298K

Pressure for carbon dioxide= 3.90*10-4atm

Temperature for carbon dioxide = 1.2*103 C0 = 1473K

Volume of carbon dioxide produce= ?

SOLUTION:

MOLES OF BUTANE REACTED AT GIVEN CONDITIONS

n= PV/RT

  = (1.50) (5.60)/(0.0821) (298)

  = 0.3433mol of butane burns

MOLES OF CARBON DIOXIDE

2mol of butane gives 8mol of carbon dioxide

0.3433 mol gives (8mol of carbon dioxide*0.3433)/2

  = 1.3733 mol of carbon dioxide

VOLUME OF CARBON DIOXIDE

V= nRT/P

  = (1.3733) (0.0821) (1473)/3.9*10-4

  = 4.2585*105L


May 19th, 2015

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