Question

Який об'єм кисню витрачається на спалювання метану, основного складника природного газу, об'ємом 25 л?

Answer 1

Ответ:

1,12. х

CH4+2O2-->CO2+2H2O

1. 2

Объяснение:

n(CH4)=V(CH4):Vm=25:22,4=1,12 моль

х=1,12×2:1=2,24моль

V(O2)=n(O2)×Vm=2,24×22,4=50,176литров

Ответ:50,176 литров

Answer 2

Oxygen Volume Required for Burning Methane

When burning methane, the stoichiometric ratio of oxygen to methane is 2:1. This means that for every 2 volumes of oxygen, 1 volume of methane is required for complete combustion.

Given the volume of methane as 25 L, we can calculate the volume of oxygen required for its combustion.

Calculation

Using the stoichiometric ratio, the volume of oxygen required can be calculated as follows: Volume of oxygen = (Volume of methane) x (2/1) Volume of oxygen = 25 L x 2 Volume of oxygen = 50 L

Therefore, the volume of oxygen required for the combustion of 25 L of methane is 50 liters.