Assume That the Combustion of Coal Is Approximated by the Burning

Question

Examlex · Accepted Answer

The Answer is By the reaction

\mathrm { C } + \mathrm { O } _ { 2 } ightarrow \mathrm { CO } _ { 2 }

then 1 mole of carbon

( 12 \mathrm {~g} )

produces 1 mole

( 44 \mathrm {~g} )

of

\mathrm { CO } _ { 2 }

. Coal produces

32.8 \mathrm { MJ } / \mathrm { kg }

and converting to

\mathrm { MWh }

using

1 \mathrm { MWh } = 3.6 \mathrm { GJ }

then

1 \mathrm { MWh }

of electricity at generation of

40 \%

efficiency requires

\frac { 1 } { 0.4 } \cdot \frac { 3.6 imes 10 ^ { 9 } \mathrm {~J} } { 3.28 imes 10 ^ { 7 } \mathrm {~J} / \mathrm { kg } } = 274 \mathrm {~kg} ext { of coal }

Since 1 mole (12

\mathrm { g } )

of carbon produces 1 mole (

44 \mathrm {~g} )

of

\mathrm { CO } 2

then

1 \mathrm {~kg}

od coal produces

( 1 \mathrm {~kg} ) imes ( 44 \mathrm {~g} / \mathrm { mol } ) / ( 12 \mathrm {~g} / \mathrm { mol } ) = 3.67 \mathrm {~kg}

of

\mathrm { CO } _ { 2 }

, then

274 \mathrm {~kg}

coal will produce

( 274 \mathrm {~kg} ) imes ( 3.67 \mathrm {~kg} / \mathrm { kg } ) = 1007 \mathrm {~kg} \mathrm { CO } { } _ { 2 }

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