Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)
In this process 3.14 moles of H₂ will be consumed.
The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-
2H₂ (g) + O₂ (g) = 2H₂O (l).
Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.
So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.
The reaction is:
4 PCl3 (g) ---> P4(s) + 6 Cl2(g).
Now, you need to convert the mass of PCl3 into number of moles, for which you use the molar mass of PCl3 in this way:
number of moles = number of grams / molar mass =>
number of moles of PCl3 = 612 g / 137.32 g/mol = 4.4567 moles of PCl3.
Now use the proportion with the ΔH rxn given.
4 mol PCl3 / 1207 kJ = 4.4567 mol / x => x = 4.4567 mol * 1207 kJ / 4 mol = 1,344.8 kJ = 1.34 * 10^3 kJ.
Answer: 1.34 * 10 ^3 kJ (option d)
