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When butane burns completely, only water and carbon dioxide gas are produced. If 11.6 g of butane and 40.0 L of oxygen at 22.0o C and 102 kPa react, what volume of carbon dioxide gas also at 22.0o C and 102 kPa can be collected over water. The vapour pressure of water at 22.0o C is 2.24 kPa.

2C4H10(g) + 13O2(g) --> 10H2O(g) + 8CO2(g)

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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