What is the balanced chemical equation for the complete combustion of ethylene? \[ C \] \[ _{2} \] \[ H \] \[ _{4} \] + 3 \[ O \] \[ _{2} \] = 2 \[ CO \] \[ _{2} \] + 2 \[ H \] \[ _{2} \] \[ O \] \[ C \] \[ _{2} \] \[ H \] \[ _{4} \] + 4 \[ O \] \[ _{2} \] = 2 \[ CO \] \[ _{2} \] + 2 \[ H \] \[ _{2} \] \[ O \] \[ C \] \[ _{2} \] \[ H \] \[ _{6} \] + 3 \[ O \] \[ _{2} \] = 2 \[ CO \] \[ _{2} \] + 3 \[ H \] \[ _{2} \] O \[ C \] \[ _{2} \] \[ H \] \[ _{4} \] + 2 \[ O \] \[ _{2} \] = 2 \[ CO \] \[ _{2} \] + 2 \[ H \] \[ _{2} \] \[ O \]

\[ C \] \[ _{2} \] \[ H \] \[ _{4} \] + 3 \[ O \] \[ _{2} \] = 2 \[ CO \] \[ _{2} \] + 2 \[ H \] \[ _{2} \] \[ O \]

What does a 50% excess air imply for ethylene combustion? 1.5 times stoichiometric air is used 50% additional fuel is needed 50% more nitrogen is present Air supply is halved

1.5 times stoichiometric air is used

Maximize profitability while ensuring efficiency, safety, and sustainability

Designing equipment and flow diagrams

Materials entering are balanced with those leaving

Regulate temperature, pressure, and flow rates

Selecting the right reactor type

Separate components based on physical or chemical properties

Minimize costs and maximize efficiency

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Process Engineering

Solve the following problems and show the complete solution including COMPLETE DIAGRAMS

Process Engineering

Process Engineering Fundamentals and Applications