Answer:
Molecular formula: C₂H₄O₂
Empirical formula: CH₂O
Explanation:
40 % C, 6.72 % H and 53.29 % O states the centesimal composition of the compound. These data means that in 100 g of compound we have x grams of a determined element.
We divide the mass by the molar mass of each:
40 g / 12 g/mol = 3.33 moles of C
6.72 g / 1 g/mol = 6.72 moles of H
53.29 g / 16 g/mol = 3.33 moles of O
We can determine rules of three to get, the molecular formula.
In 100 g of compound we have 3.33 moles of C, 6.72 moles of H and 3.33 moles of O; therefore in 60 g (1 mol) we must have
- (60 . 3.33) / 100 = 2 moles of C
- (60 . 6.72) / 100 = 4 moles of H
- (60 . 3.33) / 100 = 2 moles of O
Molecular formula is C₂H₄O₂
Empirical formula has the lowest suscripts; we divide by two, so the empirical formula is CH₂O
Below is the list of enthalpies of formation of the compounds involved in the chemical reaction.
HCl: -92.3 kJ/mol
H2: 0
Cl2: 0
Both H2 and Cl2 do not have enthaply of formation because they are pure substances.
Since, in the reaction there are 2 moles of HCl,
Enthalphy of formation = enthalpy of products - enthalphy of reactant/s
= 0 - (-92.3)(2) = +184.6 kJ
Hence, the answer is the first choice.
Acids:
Sulfuric acid, hydrochloric acid, nitric acid,
Bases:
Potassium hydroxide (KOH)
Sodium hydroxide (NaOH)
Barium hydroxide (Ba(OH)2)
