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

-262.58 kJ

Step-by-step 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=(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.