To illustrate clearly, I will rewrite the reaction in a more understandable manner.
<span>2 Al(s) + Fe</span>₂O₃ (s) ⇒ 2 Fe(s) + Al₂O₃(s) Δ<span>hrxn = –850 kJ
This reaction has a negative sign for the change in enthalpy of reaction. The sign convention only means that the reaction releases energy to the surroundings. In other words, the reaction is exothermic. Focusing on only its magnitude, this means that 850 kJ of energy is needed for this reaction of 2 Aluminum moles and 1 mole of </span>Fe₂O₃ to occur.
Now, if you only had an energy of 725 kJ, then the reaction is incomplete but it will still form Iron (Fe). We use stoichiometric calculations as follows:
725 kJ * (2 mol Fe/850 kJ) = 1.7 moles of Fe
Knowing that the molar mass of Fe is 55.6 g/mol, then the mass of produced iron is
1.7 mol Fe * 55.6 g/mol = 94.85 g iron
Answer: concentration c = 25.2 g /(41.99 g/mol × 0.75 l)
Explanation: M(NaF) = 41.99 g/mol and amount of substance n = m/M.
Concentration c = n/V
