Answer:
Hey,
QUESTION)
1) We balance the equation of the reaction, that to say :
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
2) Calculating the molar mass of simple butane :
M(C₄H₁₀) = 4M(C) + 10M(H)
- M(C₄H₁₀) = 4 x 12,0 + 10,0
- M(C₄H₁₀) = 58,0 g/mol
3) Let us calculate the quantity of material corresponding to the mass of simple butane :
n(C₄H₁₀) = m/M
- n(C₄H₁₀) = 2.46/58,0
- n(C₄H₁₀) = 4,24 x 10⁻² mol
4) Determining the progress maximal x of the reaction,
xmax = n(C₄H₁₀)/2
- xmax = 4,24 x 10⁻²/2
- xmax = 2,12 x 10⁻² mol
5) If the reagents are supplied in a stoichiometric quantity, and the combustion is complete, then :
n(CO₂) = xmax x 8
- n(CO₂) = 2,12 x 10⁻² x 8
- n(CO₂) = 17,0 x 10⁻² mol
m(CO₂) = M(CO₂) x n(CO₂)
- m(CO₂) = (12 + 2 x 16) x 17,0 x 10⁻²
- m(CO₂) = 7,48 g
The mass of carbon dioxide is therefore 7.48 g.
