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polet [3.4K]
3 years ago
15

The combustion of 14 grams of CO, according to the reaction CO(g) + ½O2(g) → CO2(g) + 67.6 kcal, gives off how much heat?

Chemistry
2 answers:
kifflom [539]3 years ago
8 0

<u>Answer:</u> The amount of heat released by the combustion of 14 g of CO is 33.8 kCal

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of CO = 14 g

Molar mass of CO = 28 g/mol

Putting values in above equation, we get:

\text{Moles of CO}=\frac{14g}{28g/mol}=0.5mol

The chemical reaction for the combustion of CO follows the equation:

CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)+67.6kCcal

By Stoichiometry of the reaction:

If 1 mole of CO produces 67.6 kCal of heat

Then, 0.5 moles of CO will produce = \frac{67.6kCal}{1mol}\times 0.5mol=33.8kCal of heat.

Hence, the amount of heat released by the combustion of 14 g of CO is 33.8 kCal

motikmotik3 years ago
3 0
The total heat is 33.8 kcal.

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If we take 2.2 grams of CO2, 6.02 X 1021 atoms of nitrogen and 0.03 gram atoms of sulphur , then the molar ratio of C, N, and O
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Answer: -

5 : 1 : 3

Explanation: -

Mass of CO₂ = 2.2 g

Molar mass of CO₂ = 44 g / mol

Number of moles of CO₂ = \frac{Mass}{Molar mass}

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Number of moles of nitrogen = \frac{Number of atoms of nitrogen}{Avogadro's number}

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Thus the mole ratio of CO₂ : N : S = 0.05 : 0.01 : 0.03

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7 0
4 years ago
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