Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l) At 1.00 atm and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 1.40 g of butane?

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

0.88m hope i am correct : )