What volume of oxygen at s.t.p is required to burn completely 7.5 dm\(^3\) of methane according to the following equation?
CH\(_4\) \(_g\) + 20\(_2\) \(_g\) → CO\(_2\) \(_g\) + H\(_2\)O\(_g\)
- 3.75 dm\(^3\)
- 7.59 dm\(^3\)
-
15.0 dm\(^3\)
- 30.0dm\(^3\)
The correct answer is: C
Explanation
1 mole of methane reacts with 2 moles of oxygen
1 vol of methane reacts with 2 vol of oxygen
: 7.5 vol of methane will yield 15 vol of oxygen in dm\(^3\)