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tamaranim1 [39]
3 years ago
8

Write the balanced chemical equation for the combustion of ethane, , and answer these questions. (Use the lowest possible coeffi

cients. Omit states of matter.) How many molecules of oxygen would combine with 16 molecules of ethane in this reaction
Chemistry
1 answer:
horrorfan [7]3 years ago
3 0

Answer:

1. 2C2H6 + 7O2 —> 4CO2 + 6H2O

2. 56moles of O2

Explanation:

Ethane undergo Combustion to produce CO2 and H20 according the equation below:

C2H6 + O2 —> CO2 + H2O

Let us balance the equation. There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 3 in front of H2O as shown below:

C2H6 + O2 —> CO2 + 3H2O

There are 2 atoms of C on left and 1atom on the right. It can be balanced by putting 2 in front of CO2 as shown below:

C2H6 + O2 —> 2CO2 + 3H2O

There are a total of 7 atoms of O on the right and 2 atoms on the left. It can be balanced by putting 7/2 in front of O2 as shown below:

C2H6 + 7/2O2 —> 2CO2 + 3H2O

Now we multiply through by 2 to remove the fraction as shown below

2C2H6 + 7O2 —> 4CO2 + 6H2O

Now the equation is balanced

2. 2C2H6 + 7O2 —> 4CO2 + 6H2O

From the equation above,

2 moles of ethane(C2H6) combined with 7moles of O2.

Therefore, 16moles of ethane(C2H6) will combine with = (16x7)/2 = 56moles of O2

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Explanation:

Let us consider the complete redox reaction:

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This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

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Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

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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
Angelina_Jolie [31]

19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.

<h3>What is vapour pressure?</h3>

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 O_2 (V) = 40 liter

Temperature (T) = 22°C = 22 + 273 = 295 K

Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm

Moles of O_2 (n) can be calculated by ideal gas equation.

PV = nRT

n = 1.007 40 ÷ 0.0821 295 = 1.663

Balanced chemical reaction;

2C_4H_10 + 13O_2 ---> 8CO_2 + 10H_2O

From reaction;

13 moles O_2 require 2 moles C_4H_10

So, 1.663 moles O_2 will require = 2 x 1.663 ÷13 = 0.256 moles of C_4H_10

Thus C_4H_10 is a limiting reagent. So it will drive the yield of CO_2.

Moles of CO_2 produced = (8/2) 0.2 = 0.8 moles

Pressure of CO_2 (P) = 102 - 2.24 = 99.76 kPa = 99.76  ÷ 101.325 = 0.985 atm

Applying the ideal gas equation for CO_2,

PV = nRT

0.985 V = 0.8 0.0821 x 295

V = 19.7 liter

The volume of CO_2 produced = 19.7 liter.

Learn more about the vapour pressure here:

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