Question

How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxygen (under constant pressure and temperature) to form carbon dioxide gas and water vapor? Assume ideal gas behavior.

Answer 1

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

$36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})$

= 126 L $O_2$

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

$105.0LO_2(\frac{4LCO_2}{7L O_2})$

= 60.0 L $CO_2$

Similarly, let's calculate the volume of water vapors formed:

$105.0L O_2(\frac{6L H_2O}{7L O_2})$

= 90.0 L $H_2O$

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

$105.0LO_2(\frac{2LC_2H_6}{7LO_2})$

= 30.0 L $C_2H_6$

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

Answer 2

Assuming ideal gas behavior, 156.128 L of gases will be in the closed reaction flask.

The equation of the reaction is;

2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g)

Number of moles of C2H6 is obtained from;

1 mole of  C2H6 occupies 22.4 L

x moles of C2H6 occupies 36.0L

x = 1.61 moles

Number of moles of O2 is obtained from;

1 mole of O2 occupies 22.4 L

x moles of O2 occupies 105.0L

x = 4.69 moles

The limiting reactant is obtained as follows;

2 moles of C2H6 reacts with 7 moles of O2

x moles of C2H6 reacts with  4.69 moles of O2

x = 1.34 moles

There is more than enough C2H6 hence O2 is the limiting reactant.

Amount of C2H6 unreacted = 1.61 moles  -  1.34 moles = 0.27 moles

Volume of unreacted C2H6 =  0.27 moles × 22.4 L = 6.048 L

Using the limiting reactant;

7 moles of O2 produces 4 moles of CO2

4.69 moles moles of O2 produces x moles of CO2

x = 2.68 moles of CO2

Volume of CO2 produced =  2.68 moles  × 22.4 L =60.032 L

For water

7 moles of O2 produces 6  moles of water

4.69 moles moles of O2 produces x moles of water

x =4.02 moles of water

Volume of water produced = 4.02 moles  × 22.4 L = 90.048 L

Total volume present after reaction = volume of unreacted C2H6 + volume of CO2 + volume of gaseous water

Total volume present after reaction = 6.048 L + 60.032 L + 90.048 L

= 156.128 L

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