Answer:
the volume of carbon dioxide formed by the combustion of 1.40 g of butane is 2.3447 L
Explanation:
Given:
2C₄H₁₀(g) + 13O₂(g) ⇒ 8CO₂(g) + 10H₂O(l)
pressure(P) = 1.00 atm
temperature(t) = 23°C
Mass of butane(m) = 1.40 g
From the balanced equation of the complete combustion of butane(C₄H₁₀), 2 moles of butane produce 8 moles of carbon dioxide(CO₂). Therefore 1 mole of butane would produce 2 mole of carbon dioxide
Therefore The molar mass of butane(M) = M(C₄H₁₀) = (12 ×4) + (1 × 10) = 48 + 10 = 58 g/mol
The amount of mole of butane(n) = mass of butane(m)/molar mass of butane(M)
n(C₄H₁₀) = m/M
n(C₄H₁₀) = 1.4/58 = 0.0241 mole
since 1 mole of butane gives 4 mole of carbon dioxide, therefore 0.0241 mole of butane = 4 × 0.0241 mole of carbon dioxide.
n(CO₂) = 4 × n(C₄H₁₀) = 4 × 0.0241 = 0.0966 mole
volume of carbon dioxide formed by the combustion of 1.40 g of butane (V) is given by
PV = nRT
∴ V = nRT/P
T = 23°C = 23 + 273 = 296 K
R = 0.08205 L atm mol⁻¹ K⁻¹
since V = nRT/P
substituting values,
V = (0.0966 × 0.0820 × 296)/1 = 2.3447
V = 2.3447 L
