atm 13. Propone (GsHo) burns in oxygen to produce carbon dioxide gas and water v

Question

atm 13. Propone (GsHo) burns in oxygen to produce carbon dioxide gas and water vapor as shown below Calculate the pressure of CO2 from 0.80 g of propane (CaHe: MM 44 g/mol) measured at 27'C in 1.2L CsHs (g) + 502 (g) -. ???? (g) + 4H2O (g) mm Hg 14. Use the Van der Waal's gas equation to calculate the pressure of a gas exerted by 2.0 moles of CO2 confined in a volume of 5.00 L at 177 °C. atm Gas momol He 0.034 0.0237 Ne o.211 0.0171 Ar 1.34 0.0322 Kr 2.32 0.0398 Xe 4.19 .0266 H2 0.244 0.0266 N 139 0.0391 02 1.36 0.0318 Ch 6.49 0.0562 CO2 3.59 0.0427 CH4 2.25 0.0428 CCI4 20.4 0.138 NH 4.17 0.037 HO 5.46 0030

Explanation / Answer

13.

mole of propane = 0.8 g/44 g/mol = 0.02 moles

mole CO2 produced = 3 x 0.02 = 0.06 moles

Using,

P = nRT/V

with,

n = 0.06 moles

R = gas constant

T = 27 oC + 273 = 300 K

V = 1.2 L

we get,

pressure of CO2 (P) = 0.06 x 0.08206 x 300/1.2 = 6154.5 atm

pressure of CO2 (P) = 6154.5 atm x 760 = 4.7 x 10^6 mmHg

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14. Van der Waals' equations,

P = [nRT/(V-nb)] - (n^2.a/V)

with,

n = 2 mol CO2

a and b taken from table above

V = 5.0 L

T = 177 oC + 273 = 450 K

R = gas constant

P = pressure of CO2

So,

P = [2 x 0.08205 x 450/(5 - 2 x 0.0427)] - (2^2 x 3.59/5)

   = 12.154 atm

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