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RUDIKE [14]
3 years ago
5

What is the change in enthalpy associated with the combustion of 530 g of methane (CH4)?Report your answer in scientific notatio

n.Your answer should have two significant figures.Use −890.8kJmol for the molar heat of combustion of methane.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

Explanation:

CH_4+2O_2\rightarrow CO_2+2H_2O,\Delta H_{comb}=-890.8 kJ/mol

Mass of methane burnt = 530 g

Moles of methane burnt = \frac{530 g}{16 g/mol}=33.125 mol

Energy released on combustion of 1 mole of methane = -890.8 kJ/mol

Energy released on combustion of 33.125 moles of methane :

-890.8 kJ/mol\times 33.125 mol=-29,507.75 kJ=-2.950775\times 10^4 kJ

-2.950775\times 10^4 kJ\approx -3.0\times 10^4 kJ

-3.0\times 10^4 kJ is the change in enthalpy associated with the combustion of 530 g of methane.

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Explanation:

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Chemical species which are capable of donating electron pairs or accepting protons are called Lewis base.

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Chemical species which are capable of donating H+ are called Bronsted acids.

Chemical species which are capable of accepting H+ are called Bronsted bases.

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2C4 H10 + 13O2 -→ 8CO2 + 10H2O If 356 moles of O2 are available to react, how many moles of CO2 will fom?
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Answer:

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The chemical equation given is:

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In this case, they are: 2 mol C₄H₁₀ :  13 mol O₂ : 8 mol CO₂ : 10 mol H₂O

Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:

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