Problem 2

A mixture of a gaseous hydrocarbon and oxygen is placed in a 1 dm

^3

vessel at 406.5 K and 101 325 Pa. The oxygen present is twice the stoichiometric amount required for combustion. After combustion, the pressure in the vessel at the same temperature increases by 5 %. Determine the molecular formula of the hydrocarbon if the mass of water formed was 0.162 g. (

R = 8.314

J mol

^{-1}

^{-1}

)

Step 3 of 4: System of mole balances

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n_1 = n(\ce{C_xH_y}) + 2\left(x + \frac{y}{4}\right)n(\ce{C_xH_y}) = 0.0300; \quad n_2 = x\,n + \left(x + \frac{y}{4}\right)n + \frac{y}{2}\,n = 0.0315

Analysis

Initial: $\ce{C_xH_y}$ plus twice stoichiometric O2 gives $n(1 + 2x + y/2) = 0.0300$ . Final: $x\,n$ of CO2 + $(x + y/4)n$ unreacted O2 + $0.009$ steam gives $n(2x + y/4) = 0.0225$ . Subtracting gives $n(1 + y/4) = 0.0075$ .