Answer:
ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol
Explanation:
Let's consider the combustion of C₈H₁₈.
C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = − 5113.3 kJ
We can calculate the standard enthalpy of formation of C₈H₁₈(g) using the following expression.
ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₈H₁₈(g)) - 25/2 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 25/2 mol × ΔH°f(O₂(g)) - ΔH°rxn
1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × (-393.5 kJ/mol) + 9 mol × (-241.8 kJ/mol) - 25/2 mol × 0 kJ/mol - (− 5113.3 kJ)
ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol
Since X is 1 g, therefore O must be 0.1 g. Therefore:
moles O = 0.1 g / (16 g / mol) = 0.00625 mol
We can see that for every 3 moles of O, there are 2 moles
of X, therefore:
moles X = 0.00625 mol O (3 moles X / 2 moles O) =
0.009375 mol
Molar mass X = 1 g / 0.009375 mol
<span>Molar mass X = 106.67 g/mol</span>
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