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Sagot :
19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.
What is vapour pressure?
Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.
Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2
Volume of [tex]O_2[/tex] (V) = 40 liter
Temperature (T) = 22°C = 22 + 273 = 295 K
Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm
Moles of [tex]O_2[/tex] (n) can be calculated by ideal gas equation.
PV = nRT
n = 1.007 40 ÷ 0.0821 295 = 1.663
Balanced chemical reaction;
2[tex]C_4H_10[/tex] + 13[tex]O_2[/tex] ---> 8[tex]CO_2[/tex] + 10[tex]H_2O[/tex]
From reaction;
13 moles [tex]O_2[/tex] require 2 moles [tex]C_4H_10[/tex]
So, 1.663 moles [tex]O_2[/tex] will require = 2 x 1.663 ÷13 = 0.256 moles of [tex]C_4H_10[/tex]
Thus [tex]C_4H_10[/tex] is a limiting reagent. So it will drive the yield of [tex]CO_2[/tex].
Moles of [tex]CO_2[/tex] produced = (8/2) 0.2 = 0.8 moles
Pressure of [tex]CO_2[/tex] (P) = 102 - 2.24 = 99.76 kPa = 99.76 ÷ 101.325 = 0.985 atm
Applying the ideal gas equation for [tex]CO_2[/tex],
PV = nRT
0.985 V = 0.8 0.0821 x 295
V = 19.7 liter
The volume of [tex]CO_2[/tex] produced = 19.7 liter.
Learn more about the vapour pressure here:
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