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The combustion of 45.0 g of methane (natural gas) releases 2498 kJ of heat energy. What amount of energy (in kilocalories) would be released for the combustion of 0.440 oz (ounces) of methane?

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Final answer:

The combustion of 0.440 oz of methane would release approximately 165.331 kilocalories of energy, after converting the mass from ounces to grams and using the given data for the combustion of methane.

Step-by-step explanation:

To calculate the amount of energy released for the combustion of 0.440 oz of methane, we first need to convert ounces to grams. There are 28.3495 grams in an ounce, so 0.440 oz is equivalent to 0.440 × 28.3495 g = 12.4736 grams of methane.

Now we look at the given information: the combustion of 45.0 g of methane releases 2498 kJ of heat energy. We need to find out how much energy is released from 12.4736 g of methane by setting up a proportion:

· 45.0 g CH₄ ----- 2498 kJ
· 12.4736 g CH₄ ----- x kJ

Using the cross-multiplication method, we find x:

x = (2498 kJ × 12.4736 g) / 45.0 g = 691.685 kJ

To convert kilojoules to kilocalories, we use the conversion factor 1 kcal = 4.184 kJ:

x = 691.685 kJ × (1 kcal / 4.184 kJ) = 165.331 kcal

Therefore, the combustion of 0.440 oz of methane would release approximately 165.331 kilocalories of energy.

User Anshulkatta
by
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4 votes

Answer: 166.13 kcal

Explanation:

1 ounce = 28.35 g

0.440 ounce=
(28.35)/(1)* 0.440=12.47g

Given: 45.0 g of methane releases= 2498 kJ of heat energy

Thus 12.47 g of methane releases=
frac {2498}{45}* 12.47=692.2kJ of heat energy

1 kJ =0.24 kcal

692.2 kJ=
\frac {0.24}{1}* 692.2=166.13kcal

User Adam Hollidge
by
5.4k points