Question
Asked Nov 25, 2019
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Given the following reaction, what volume of oxygen would be needed to react with 250. grams of propane at 551 °C and 0.976 atm?
C3H8 (g) + 5 O2 (g) -> 3 CO2 (g) + 4 H2O (g)

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Expert Answer

Step 1

Given information,

ЗСО)
4 H O
C,H 502
and O2 is 1:5.
From the eqn. themole ratiobetweenC3H
Mass of CH250g; Temperature = 551°C and Pressure = 0.976 atm
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ЗСО) 4 H O C,H 502 and O2 is 1:5. From the eqn. themole ratiobetweenC3H Mass of CH250g; Temperature = 551°C and Pressure = 0.976 atm

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Step 2

Calculate the volume...

250 g
mol CHs
= 5.6689 mol
44.1g/mol
Fromideal gasequation
atm L
(824.15K
mol K
(5.6689mol 0.082 1-
nRT
V =
0.976atm
383.5732
L 393L
0.976
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250 g mol CHs = 5.6689 mol 44.1g/mol Fromideal gasequation atm L (824.15K mol K (5.6689mol 0.082 1- nRT V = 0.976atm 383.5732 L 393L 0.976

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