Final answer:
To find the heat produced by the reaction of 38.1 g of methanol with excess oxygen, first calculate the moles of methanol, then use the stoichiometry of the reaction to find the heat. A total of 1059.1 kJ of heat is produced.
Step-by-step explanation:
To calculate the amount of heat produced when 38.1 g of methanol reacts with 59.1 g of oxygen, we need the balanced thermochemical equation for the combustion of methanol. The provided information includes a related reaction of methanol with oxygen:
CH₃OH(g) + 1.5O₂(g) → CO₂(g) + 2H₂O(l) ΔH = -890 kJ/mol
First, we determine the moles of methanol:
- Molar mass of CH₃OH = 32.04 g/mol
- Moles of CH₃OH = 38.1 g / 32.04 g/mol = 1.19 moles
Then, we use the stoichiometry of the combustion reaction to relate the moles of methanol to heat produced:
- Heat produced by 1.19 moles of CH₃OH = 1.19 moles × 890 kJ/mol = 1059.1 kJ
The weight of oxygen is not directly used in the calculation, provided that there is excess oxygen. Finally, the heat produced by the reaction is 1059.1 kJ.