[Bronx19]

Any assistance would be appreciated.

Using the following data calculate the heat of reaction for the production of coal gas: 2 C(s) + 2 H2O(g)  CH4(g) + CO2(g)

C(s) + H2O(g) = CO(g) + H2(g)    Standard Heat of reaction =+131.3 kJ/mol
CO(g) + H2O(g) = CO2(g) + H2(g)   Standard Heat of reaction = -41.2 kJ/mol
CO(g) + 3 H2(g) = CH4(g) + H2O(g)   Standard Heat of reaction = -206.1 kJ/mol

Thus far I have C = (2 x +131.3), CH4 = (-206.1) and CO2 = (-41.2). However, Im stuck on H20. Do I leave as is, since H20 is already a reactant along with C in the first equation, or do I use another?

The value of H0 for the following reaction is -126 kJ. Determine the amount of heat (in kJ) that would be evolved by the reaction of 25.0 g of Na2O2 with water.

2Na2O2(s) + 2H2O(l) = 4NaOH(s) + O2(g)

25g/77.98gmol = 0.32mol Na2O2.

0.32mol x 126kJ = 40.3kJ. Now, this is the heat for 25g/0.32mol of substance, but the stoichiometric equation states that two moles are required. Does this mean I should be halving or doubling this amount?

[iceman]

yup, you have imput the kj/mol of water too. every compound haso have a value when u calculate it

[Yggdrasil]

Any assistance would be appreciated.

Using the following data calculate the heat of reaction for the production of coal gas: 2 C(s) + 2 H2O(g)  CH4(g) + CO2(g)

C(s) + H2O(g) = CO(g) + H2(g)    Standard Heat of reaction =+131.3 kJ/mol
CO(g) + H2O(g) = CO2(g) + H2(g)   Standard Heat of reaction = -41.2 kJ/mol
CO(g) + 3 H2(g) = CH4(g) + H2O(g)   Standard Heat of reaction = -206.1 kJ/mol

Thus far I have C = (2 x +131.3), CH4 = (-206.1) and CO2 = (-41.2). However, Im stuck on H20. Do I leave as is, since H20 is already a reactant along with C in the first equation, or do I use another?

I don't quite understand the approach you are using to solve this question.  Are you trying to find the standard enthalpies of formation of each of the reactants and products from the information given?  If so, the standard enthalpy of formation of elemental carbon is zero since elemental forms are defined to have enthalpies of formation of zero.

However, an easier way, is to add and subtract the given equations until you end up with the equation you want.  If you want an example, I can try to make one up for you.

The value of H0 for the following reaction is -126 kJ. Determine the amount of heat (in kJ) that would be evolved by the reaction of 25.0 g of Na2O2 with water.

2Na2O2(s) + 2H2O(l) = 4NaOH(s) + O2(g)

25g/77.98gmol = 0.32mol Na2O2.

0.32mol x 126kJ = 40.3kJ. Now, this is the heat for 25g/0.32mol of substance, but the stoichiometric equation states that two moles are required. Does this mean I should be halving or doubling this amount?

126kJ of heat are released for every two moles of Na₂O₂ reacted.  Therefore, you need to halve the 40.3kJ to get the correct number.

(note: Na₂O₂ does not make sense as a compound.  Are you sure that the correct formula is not Na₂O?)