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Sphinxa [80]
4 years ago
12

the molar enthalpy of formation of carbondioxide is -393kjmol. calculate the heat released by the burning of 0.327g of carbon to

form carbondioxide.
Chemistry
1 answer:
Arte-miy333 [17]4 years ago
7 0

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

<u>Explanation</u>:

The standard enthalpy change of reaction,  Δ H ∘ , is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.

                                    C (s]  +  O 2(g] → CO 2(g]

Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.

First to convert  grams of carbon into moles,

use carbon's molar mass(12.011 g).

                    Moles of C = mass in gram / molar mass

                                        = 0.327 g  / 12.011 g

                     Moles of C = 0.027 moles

Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.

                                        =  0.027 moles C  \times 393 kJ

             Heat released  = 10.611  kJ.

So, when  0.027  moles of carbon react with enough oxygen gas, the reaction will give off  10.611 kJ  of heat.

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

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<h3>Further Explanation</h3>
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Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(aq) + 2H₂O(l)

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  • For example a reaction between sodium hydroxide and sulfuric acid.

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Keywords: Precipitation

<h3>Learn more about: </h3>
  • Precipitation reaction: brainly.com/question/11194650
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Level: High school

Subject: Chemistry

Topic: Chemical reactions

Sub-topic: Precipitation reactions

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