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Calcium hydroxide, which reacts with carbon dioxide to form calcium carbonate, was used by the ancient Romans as mortar in stone structures. The reaction for this process is Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(g) ΔH = –69.1 kJ What is the enthalpy change if 3.8 mol of calcium carbonate is formed?

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aimartinezm

Answer:

-262.58 kJ

Explanation:

[tex]Ca(OH)_{2}(s) + CO_{2}(g) - ->  CaCO_{3}(s) + H_{2}O(g)[/tex]

The enthalpy represents the energy that is either absorbed or released during a chemical reaction.  

We see that the chemical reaction that we have is balanced and 1 mol of calcium carbonate was formed. During this process -69.1 kJ or energy was released because it has a negative sign. We can say that the enthalpy changes  is -69.1kJ per every mol of calcium carbonate formed.  

Using a simple rule of three we can get the enthalpy change when 3.8 mol of [tex]CaCO_{3}[/tex] are formed.

-69.1 Kj --- > 1 mol of  [tex]CaCO_{3}[/tex]

X         --- >  3.8 mol of [tex]CaCO_{3}[/tex]

[tex]x=\frac{3.8 mol-of-CaCO_{3}* -69.1 kJ }{1 mol-of-CaCO_{3}} = -262.58kJ[/tex]

So, the energy released when 3.8 mol of calcium carbonate are formed is -262.58 kJ.  

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