Question

Given the following reaction: 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) +

4H₂O(g) ∆H = -1277 kJ What is the energy change when 11.8 g of
CH₃OH react with excess O₂?

Answer 1

The enthalpy change when 11.8 g of CH₃OH react with excess O₂ is approximately -234.7 kJ, calculated by first determining the moles of CH₃OH and then using the reaction's stoichiometry.

Step-by-step explanation:

The student's question pertains to the calculation of the enthalpy change of a reaction involving methanol (CH₃OH) and oxygen (O₂). First, we need to determine the amount in moles of CH₃OH that corresponds to 11.8 g. The molar mass of CH₃OH is approximately 32.04 g/mol. Thus, 11.8 g of CH₃OH is equivalent to 11.8 g / 32.04 g/mol = 0.368 moles of CH₃OH.

Using the stoichiometry of the balanced equation, 2 moles of CH₃OH react with excess O₂ to produce an enthalpy change of -1277 kJ. To find the enthalpy change for 0.368 moles of CH₃OH, we set up a proportion:

(0.368 moles CH₃OH / 2 moles CH₃OH) = x kJ / -1277 kJ

Solving for x gives us the enthalpy change for 11.8 g of CH₃OH reacting:

x = (0.368 / 2) × -1277 kJ

x ≈ -234.7 kJ

Therefore, when 11.8 g of CH₃OH react with excess O₂, the energy change is approximately -234.7 kJ.

Answer 2

The energy change when 11.8 g of CH3OH react with excess O2 is -234.5 kJ.

Step-by-step explanation:

The energy change when 11.8 g of CH3OH react with excess O2 can be calculated using the given enthalpy change (ΔH) and the stoichiometry of the reaction. The stoichiometry tells us that 2 mol of CH3OH react with 3 mol of O2 to produce 2 mol of CO2 and 4 mol of H2O. We can convert the mass of CH3OH to moles using its molar mass, and then use the ratio of moles to calculate the energy change:

Molar mass of CH3OH = 32.04 g/mol

Number of moles of CH3OH = 11.8 g / 32.04 g/mol = 0.367 mol

Energy change = ΔH * (0.367 mol / 2 mol) = -1277 kJ * 0.367 / 2 = -234.5 kJ