Answer:
65.71%
Explanation:
First, we can write the mass of the mixture, thus:
7.519g = X + Y <em>(1)</em>
<em>Where X is the mass of methane and Y the mass of butane</em>
<em />
Also, the reactions of combustion are:
CH₄ + 2O₂ → CO₂ + 2H₂O
<em>2 moles of oxygen react per mole of methane</em>
C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O
<em>13/2 moles of oxygen react per mole of methane</em>
<em />
That means, in therms of moles of oxygen we can write:
0.9050 moles = 2X/16.04 + 13/2Y/ 58.12
0.9050 = 0.12469X + 0.11184Y <em>(2)</em>
<em>Where 16.04 and 58.12 are molar masses of methane and butane</em>
That is because if X is the mass of methane:
X g Methane * (1mol / 16.04g) = Moles methane
Moles methane * (2 moles Oxygen / mole methane) = Moles oxygen
Replacing (1) in (2):
0.9050 = 0.12469X + 0.11184 (7.519 - X)
0.9050 = 0.12469X + 0.841 - 0.11184X
0.0641 = 0.01285X
X = 4.988g = Mass of methane.
And mass percent of methane is:
4.988g / 7.591g * 100
<h3>65.71%</h3>
