Naphthalene (C10H8)is a solid aromatic compound often sold as mothballs.The complete combustion of this substance to yield CO2(g)and H2O(l)at 25 ْC yields -5154 kJ/mol.

A. write balanced equation for the combustion of C10H8.
C10H8+12O2->10CO2+4H2O ΔH =-5154
B. write a balanced equation for the formation of C10H8 fro the elements and calculate its ΔH o f.
5154= [x+0]-[10(-394)+4(-286)]
5154=[x+0]-[-3940+-1144]
5154=x+5084
x=70

I think you have the correct answer but I'm not sure you arrived at it correctly.

Isn't
dHrxn= dHproducts - dHreactants.
That will give you 70 kJ/mol for C10H8.
You have worked it as
reactants - products and changed the sign of dH combustion.

Napthalene has the formula C10H8.It burns in air according to the equation

C10H8+12O----->10Co2+4H2O
16g of napthalene was burnt in excess oxygen.
a)Calculate the number of moles of napthalene burnt.
b)Calculate the mass of water produced in the combustion .
c)Calculate the volume of Co2 produced at r.t.p.
d)Clculate the moles of oxygen needed for the reaction.

A. The balanced equation for the combustion of C10H8 is:

C10H8 + 12O2 -> 10CO2 + 4H2O

B. To calculate the enthalpy change (ΔH) for the formation of C10H8 from the elements, we need to subtract the enthalpy of formation of the products from the enthalpy of formation of the reactants.

The enthalpy of formation of CO2 is -394 kJ/mol, and the enthalpy of formation of H2O is -286 kJ/mol.

ΔHf = [ΔHf of products] - [ΔHf of reactants]

ΔHf = [0] - [10(-394) + 4(-286)]
= 0 - [-3940 - 1144]
= 0 - (-5084)
= 5084 kJ/mol

Therefore, the enthalpy change (ΔH) for the formation of C10H8 from the elements is 5084 kJ/mol.

A. The balanced equation for the combustion of C10H8 (naphthalene) is:

C10H8 + 12O2 -> 10CO2 + 4H2O

B. To determine the ΔHof (standard enthalpy of formation) of C10H8, we need to calculate the energy change involved in forming one mole of C10H8 from its elements in their standard states. The elements in their standard states are carbon in the form of graphite (C), hydrogen gas (H2), and oxygen gas (O2).

The balanced equation for the formation of C10H8 from its elements is:

10C + 4H2 + ΔHof = C10H8

From the given information, we know that the ΔH of the combustion reaction is -5154 kJ/mol. Using the fact that the formation reaction is the reverse of the combustion reaction, the ΔHof for the formation of C10H8 can be determined as:

ΔHof = -ΔHcombustion
= -(-5154 kJ/mol)
= 5154 kJ/mol

Now, let's calculate the value of x in your equation:

5154 = [x + 0] - [10(-394) + 4(-286)]
5154 = x + 0 - [-3940 - 1144]
5154 = x + 5084
x = 70

Therefore, the ΔHof for the formation of C10H8 is 70 kJ/mol.