Given data:
Feed to the reactor is,
Methane = 60 mole %
Oxygen = 40 mole %
Fractional conversion of methane = 0.622
Selectivity = 4.1 mol HCHO/ mol CO_{2}
Formaldehyde is produced by the oxidation of methane in presence of vanadium oxide. Carbon dioxide also produced in an undesired side reaction.
The reactions are given as follows:
${\mathrm{CH}}_{4}+{\mathrm{O}}_{2}\to \mathrm{HCHO}+{\mathrm{H}}_{2}\mathrm{O}$ desired reaction
${\mathrm{CH}}_{4}+2{\mathrm{O}}_{2}\to {\mathrm{CO}}_{2}+2{\mathrm{H}}_{2}\mathrm{O}$ undesired reaction
Basis: 100 mol fed to the reactor.
Therefore, the number of moles of methane and oxygen fed to the reactor becomes,
CH_{4} = 60 mol
Oxygen = 40 mol
The fraction la conversion of methane = 0.662
Therefore, the total number of moles of methane reacted becomes,
$\begin{array}{rcl}{\mathrm{CH}}_{4}\left(\mathrm{reacted}\right)& =& 60\mathrm{mol}\times 0.662\\ & =& 39.72\mathrm{mol}\end{array}$
The number of moles of methane required producing formaldehyde and carbon oxide is not given but selectivity of formaldehyde per carbon dioxide is given.
Selectivity of formaldehyde per mol of carbon dioxide = 4.1 mol HCHO / mol CO_{2}
Assume x mole of methane reacts in desired reaction and 39.72 – x mol of methane reacts in undesired reaction.
The whole scenario can be represented on a diagram as follows:
From the above diagram,
mol HCHO / mol CO_{2} = x/39.72-x ………. (1)
From the selectivity data,
mol HCHO / mol CO_{2} = 4.1 mol HCHO /mol CO_{2 …………. (2)}
Equate equation (1) and (2)
$\begin{array}{rcl}\frac{\mathrm{x}}{39.72-\mathrm{x}}& =& \frac{4.1}{1}\\ \mathrm{x}& =& 4.1\times \left(39.72-\mathrm{x}\right)\\ \mathrm{x}& =& 162.852-4.1\mathrm{x}\\ 5.1\mathrm{x}& =& 162.852\\ \mathrm{x}& =& 31.93\mathrm{mol}\end{array}$
The number of moles of methane reacts in desires reaction = 31.93 mol
Therefore, the number of moles of methane reacts in undesired reaction becomes = 39.72 – 31.93
= 7.79 mol
Fractional yield of a chemical compound is the ratio of moles of that compound formed to the number of moles of reactant consumed.
For the production of HCHO, the formula can be written as follows:
$\mathrm{Yield}=\frac{\mathrm{mol}\mathrm{of}\mathrm{HCHO}\mathrm{formed}}{\mathrm{mol}\mathrm{of}{\mathrm{CH}}_{4}\mathrm{consumed}}$ ............. (3)
mol of HCHO formed = 31.93 mol
mol of CH4 reacted = 39.72 mol
Plugin the values in equation (3)
$\begin{array}{rcl}\mathrm{Yield}& =& \frac{31.93\mathrm{mol}}{39.72\mathrm{mol}}\\ & =& 0.803\simeq 0.8\end{array}$
Fraction yield of formaldehyde production is 0.8.