Answer: -2.9 X 10^4 kJ
Explanation:
First, calculate the number of moles of CH4 using a formula mass of 16.043gmol from the periodic table.
530g × 1.0 mol /16.043g = 33.036 mol
Next, calculate the total change in enthalpy.
ΔH = 33.036 mol × −890.8kJ/mol = −29,428kJ
The answer should have two significant figures, so round to −2.9 × 10^4 kJ.
The change in enthalpy associated with the combustion of 530 g of methane is -2.94 × 10⁴ kJ.
Let's consider the thermochemical equation for the combustion of methane.
CH₄(g) + 2 O₂(g) ⇒ CO₂(g) + 2 H₂O(l) ΔH = -890.8 kJ/mol
We can calculate the change in enthalpy (ΔH) associated with the combustion of 530 g of methane using the following relationships.
[tex]\Delta H = 530 g \times \frac{1mol}{16.04g} \times \frac{(-890.8kJ)}{mol} = -2.94 \times 10^{4} kJ[/tex]
The change in enthalpy associated with the combustion of 530 g of methane is -2.94 × 10⁴ kJ.
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