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Methyl alcohol (CH3OH)(CH_3OH) burns with 200% theoretical air, yielding CO2,H2O,O2,CO_2, H_2O, O_2, and N2.N_2. Determine the a. balanced reaction equation. b. air-fuel ratio on a mass basis. c. molar analysis of the products.

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(a) The chemical equation

CH3OH+(2.0)a(O2+3.76N2)bCO2+cH2O+aO2+dN2\begin{align*} CH_3OH + (2.0)a(O_2 + 3.76N_2) \rightarrow b CO_2 + cH_2O + aO_2 + dN_2 \end{align*}

To balance the equation we have

C:1=bH:4=2cO:1+4a=2b+c+2aN:2a(3.76)(2)=2d\begin{align*} C:& \quad 1 = b \\ H:& \quad 4 = 2c \\ O:& \quad 1 + 4a = 2b + c + 2a \\ N:& \quad 2a(3.76)(2) = 2d \\ \end{align*}

Thus, b=1b = 1; c=2c = 2, a=1.5a = 1.5, d=11.28d = 11.28

The balanced reaction is

CH3OH+(2.0)(1.5)(O2+3.76N2)CO2+2H2O+1.5O2+11.28N2\begin{align*} CH_3OH + (2.0)(1.5)(O_2 + 3.76N_2) \rightarrow CO_2 + 2H_2O + 1.5O_2 + 11.28N_2 \end{align*}

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