First, we need to be sure that we have identified the correct naphtha component so that we can find its heat or enthalpy of combustion, which is required for the first part of the question and is not provided. The entry ΔHf = -204.6 kJ/mol given in the table at the end of the question refers to the heat of formation of the hydrocarbon, which is much less than the heat of combustion.
(a) the complete combustion of a 1.722 g sample of the hydrocarbon produced
- 5.278 g of carbon dioxide, equivalent to 5.278/44.01 = 0.1199 moles of CO2 and
. containing 12.01*0.119 = 1.440 g of atomic carbon
- 2.521 g of water, equivalent to 2.521/18.015 = 0.1399 moles of H2O, containing
. 2*1.008*0.1399 = 0.2821 g of atomic hydrogen.
As a check, we note that 1.440 + 0.2821 = 1.722 g - the weight of the hydrocarbon sample.
Allowing for the fact that each mole of water produced contains two atoms of hydrogen, the empirical formula of the hydrocarbon is obtained from the mole ratio
C : H = 0.1199 : 2*0.1399 = 1 : 2.333 = 3 : 7
that is, the empirical formula is C3H7. We are told that the molecular formula is twice this, and that the carbon chain is branched, which identifies the hydrocarbon as an iso-hexane C6H14, ie either 2- or 3-methyl-pentane. From the data in the table provided, it is clearly the first of these.
The Wikipedia article below gives the heat or enthalpy of combustion for this compound as -4163 kJ/mol.
(b) the heat required to warm 25.2 m³ of water, with specific heat 4.18 J/g.K, from 16.7 to 38.1°C is
Q = 25.2 x 10^6 x 4.18 x (38.1 - 16.7) = 2.254 x 10^9 J
assuming a water density of 1.00 g/mL.
Each litre of isohexane, with density 0.6532 g/mL, weighs 653.2 g, equivalent with a molar mass of 86.17 g to 7.580 moles. Upon combustion it will produce 7.580*4163 = 31556 kJ.
So that a total of 2.254 x 10^9/(31556 x 10^3) = 71.4 litres of isohexane will be required.
See Wikipedia article and data page on 'Hexane'