Question

Calculate the volume of O2 required for complete combustion of 12.5 dm3 ethane gas at NTP

Answer 1

Answer:

Hi mate

According to reaction, At STP; 1 mole of acetylene that is 22.4L require 5/2 moles of oxygen that is 5 2 × 22 . 4 = 56 L for complete combustion.

HOPE IT HELPED YOU..

Explanation:

Answer 2

Answer:

43.74 dm³

Explanation:

The combustion reaction of ethane is shown below as:

$2C_2H_6_{(g)}+7O_2_{(g)}\rightarrow 4CO_2_{(g)}+6H_2O_{(l)}$

From the reaction,

2 moles of ethane reacts with 7 moles of Oxygen gas

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 dm³ atm/ K mol  

n = PV / RT

So, at NTP, pressure = 1 atm and temperature = 293.15 K are constant. SO, moles can be replaced with volume.

Volume of 2 moles = 2 * 0.0821 * 293.15 / 1 = 48.14 dm³

Volume of 7 moles = 168.47 dm³

Thus,

48.14 dm³ of ethane reacts with 168.47 dm³ of Oxygen gas

1 dm³ of ethane reacts with 168.47/48.14 dm³ of Oxygen gas

12.5 dm³ of ethane reacts with 168.47/48.14 * 12.5  dm³ of Oxygen gas

Volume of oxygen gas = 43.74 dm³