The planet Uranus indeed contains a significant amount of hydrogen and methane, both highly flammable gases. However, the burning of methane or hydrogen requires oxygen. Simply put, there is no free oxygen on the planet Uranus. On earth, we are so immersed in oxygen that we tend to take it for granted. Many chemical reactions that require oxygen seem to just happen automatically on earth: metals rust, forests catch fire, and candles burn. We may be tempted to ignore oxygen's role in a chemical reaction since it seems to be always there. But oxygen is not always present. If I place metal in a jar containing only argon, it will not rust. If I place a lit candle in a jar with no oxygen, its flame will go out. There is an easy demonstration you can do at home to convince yourself of this fact. Get a bowl and fill it about a quarter full with baking soda and vinegar. The baking soda will react with the vinegar, releasing carbon dioxide gas that fills up the bowl and pushes the oxygen out. If you are careful to not disturb the bowl, then the carbon dioxide will stay in the bowl and keep the oxygen out since carbon dioxide is heavier than oxygen. Now, light a match and slowly move it into the bowl. The moment that the match hits the invisible oxygen-depleted pocket of carbon dioxide gas in the bowl, it will immediately go out. This simple demonstration makes the role of oxygen in combustion obvious.

Dear student

Assume that 100 g each of hydrogen, oxygen and methane are present.

The volume of gases is directly proportional to the number of moles of gases.

The molar mass of hydrogen, oxygen and methane are 2 g/mol, 32 g/mol and 16 g/mol respectively.

The number of moles of hydrogen, oxygen and methane are

2/ 100 ,

32/ 100

and

16 1/100

respectively.

The ratio of the volumes of hydrogen, oxygen and methane will be

2 1/100 : 32 /100 : 16 1/100 =

1

1

:

16

1

:

8

1

=16:1:2

Hello There,

First of all calculate the total number of moles using the ideal gas law. Here as you have not mentioned temperature I have assumed it as room temperature.

BY formula n = PV/RT you'll get total moles = 0.5

Now calculate the total mass by the formula total moles = moles of ch4 + moles of SO

I.e 0.5 = x/16 + x/48

By this, you'll get x = 0.5 g

Now calculate number of moles of individual component by formula n = 0.5 g / 16 for CH4 and 0.5/48 for SO.

From the obtained values of moles of CH4 and SO, calculate the Mole fraction

which is obtained as 0.75 for CH4 and 0.25 for SO.

After this,

We have for gases, Pressure fraction = Mole Fraction

i.e p1/12 = 0.75 for CH4 and 12-p1 for SO

The obtained values for partial pressure for CH4 is 9 atm and for SO is 3 atm

Good Luck!!!!