Question #bbfd1

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Answer 1 · 2015-04-30T21:34:38

2480.6 kJ required.

For:

$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)}$ $CH_(4(g))+2O_(2(g))rarrCO_(2(g))+2H_2O_((l))$

$DeltaH_(comb)=-882.0"kJ"$

So for the reverse reaction:

$CO_(2(g))+2H_2O_((l))rarrCH_(4(g))+2O_(2(g))$

$DeltaH=+882.0"kJ"$

$[M_r]CH_4=16.0$

So $16.0"g"$ requires $882.0"kJ"$

So $1"g"rarr882.0/16.0"kJ"$

So $45"g"rarr(882.0)/(16.0)xx45=2480.0"kJ"$