Question

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?

Answer 1

Answer:

-262.58 kJ

Explanation:

$Ca(OH)_{2}(s) + CO_{2}(g) - -> CaCO_{3}(s) + H_{2}O(g)$

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 $CaCO_{3}$ are formed.

-69.1 Kj --- > 1 mol of  $CaCO_{3}$

X         --- >  3.8 mol of $CaCO_{3}$

$x=\frac{3.8 mol-of-CaCO_{3}* -69.1 kJ }{1 mol-of-CaCO_{3}} = -262.58kJ$

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