What is the volume of oxygen required to burn complete 45 cm\(^3\) of methane at s.t.p?

talk2kb

2 years ago

Combustion reaction of Methane (Hydrocarbon) to yield CO2 and Water
balancing the equation of Methane you have
1Mole of Methane will react with
2Mole of Oxgygen i.e 45*2=90

SpeDan

2 years ago

Oxygen=32
Methane=16 at STP
Therefore 32 of O = 16 of CH4
X of O = 45 of CH4
X = 32*45/16
X = 90cm3

De sage

1 year ago

I want to ask a question my school.
what is The essence of putting
STP in the question. I solved with 22.4 dm³ since I couldn't see the option there. I had to remove it.

solving using the STP you mentioned in the question

CH4 + 2O2--> CO2+2H2O
1dm³ of CH4--> 2×22.4dm³ of O2
0.045dm³ of CH4--> x
cross multiply
x= 44.8× 0.045
x= 2.016dm³.

this is because you mentioned STP.

though I solved like this and the ANS wasn't there. I removed 22.4dm³

then do 45cm³ × 2 = 90cm³

nwankwoimmaculate

5 years ago

The equation you gave in your answer is not balanced

CBA

7 years ago

Correct

quadri101

2 months ago

dm3 made me confused 😕

Marh_xahra

1 year ago

the explanation given us not enough to understand the question

nwankwoimmaculate

5 years ago

The equation you gave in your answer is incorrect (4 atoms of oxygen reactants and three atoms of oxygen product,)

Tega204

2 years ago

CH4 + 2O2----- Co2 + 2H20

CH4: 2O2
1:2
CH4= 12+4=16g/mol
2O2=64g/mol

CH4-----2O2
16g/mol ----- 64g/mol
45cm³---- Xcm³
16X= 45 x 64
X=45 x 64/16=180cm³