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Consider the following thermochemical equation: 2CH 3 OH(l) + 3O 2 (g) → 2CO 2 (g) + 4H 2 O(g) ∆H o rxn = –1277 kJ What mass of water can be heated from 23.5°C to 88.2°C with the heat that is released from the combustion of 12.0 g of CH 3 OH (molar mass = 32.0 g/mol), assuming no heat is lost to the surroundings? Specific heat capacity of H 2 O = 4.184 J/g°C.

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Answer:

1.77 × 10³ g

Explanation:

Let's consider the following thermochemical equation:

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g) ∆H°rxn = –1277 kJ

1277 kJ of heat are released per 2 moles of methanol (molar mass = 32.0 g/mol). The heat released upon the combustion of 12.0 g of  methanol is:

[tex]12.0gCH_{3}OH.\frac{1molCH_{3}OH}{32.0gCH_{3}OH}.\frac{(-1277kJ)}{1molCH_{3}OH} =-479kJ[/tex]

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the water is zero.

Qc + Qw = 0

Qw = -Qc = - (-479kJ) = 479 kJ

For water, we know that

Qw = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in the temperature

Qw = c × m × ΔT

479 × 10³ J = (4.184 J/g.°C) × m × (88.2°C - 23.5°C)

m = 1.77 × 10³ g

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