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How much heat is evolved or absorbed in this problem?

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Limestone, CaCo3, when subjected to a temperature of 900 degrees Celsius in a kiln, decomposes to calcium oxide and carbon dioxide. How much heat is evolved or absorbed when one gram of limestone decomposes?

The standard enthalpies of formation at 25 degrees Celsius(kj/mol) of these compounds at 1 atm, Aqueous ions at 1 M is:

CaCO3 -1206.9
CaO -635.1
CO2 -393.5

Nitroglycerin is a powerful explosive, giving four different gases when detonated.2 C3H5(NO3)3 (l) → 3 N2 (g) + 1/2 O2 (g) + 6 CO2 (g) + 5 H2O (g)Given that the enthalpy of formation of nitroglycerin, ΔHf°, is −364 kJ/mol, calculate the energy (heat at constant pressure) released by this reaction.Δ= Σn􀁄rxnHfΔ􀁄H(prod.) − ΣmfΔ􀁄H(reac.)= (3 mol)(0) + (1/2 mol)(0) + (6 mol)(−393.5 kJ/mol) + (5 mol)(−241.8 kJ/mol) − (2 mol)(−364 kJ/mol) rxnΔ􀁄H= −2842 kJ rxnΔ􀁄Hb) What is delta H when 4.65 mol of products is formed?

Your question was answered above.

dHrxn = (n*dH products) - (n*dH reactants)

To determine the amount of heat evolved or absorbed during the decomposition of one gram of limestone (CaCO3), we need to calculate the change in enthalpy for the reaction.

The balanced equation for the decomposition of limestone is:
CaCO3(s) → CaO(s) + CO2(g)

To calculate the change in enthalpy (ΔH) for this reaction, we can use the enthalpies of formation of the products and reactants.

ΔH = Σ(n * ΔH_f(products)) - Σ(n * ΔH_f(reactants))

Where:
ΔH = change in enthalpy
Σ = sum
n = stoichiometric coefficients
ΔH_f = enthalpy of formation

In this case, the stoichiometric coefficients for all the compounds are 1.

Let's calculate the change in enthalpy (ΔH) using the given enthalpies of formation:

ΔH = (1 * ΔH_f(CaO)) + (1 * ΔH_f(CO2)) - (1 * ΔH_f(CaCO3))

ΔH = (1 * -635.1 kJ/mol) + (1 * -393.5 kJ/mol) - (1 * -1206.9 kJ/mol)

ΔH = -635.1 kJ/mol - 393.5 kJ/mol +1206.9 kJ/mol

ΔH = 178 kJ/mol

So, for the decomposition of one mole of limestone, 178 kJ of heat is evolved.

To find the heat evolved or absorbed when one gram of limestone decomposes, we need to convert the moles of limestone to grams.

The molar mass of CaCO3 is 100.09 g/mol.

Therefore, the number of moles of CaCO3 in one gram is:

moles = mass / molar mass
moles = 1 g / 100.09 g/mol
moles = 0.00999 mol

To find the heat evolved for one gram of CaCO3, we can calculate:

heat evolved = ΔH * moles
heat evolved = 178 kJ/mol * 0.00999 mol

heat evolved = 1.78 kJ

Hence, when one gram of limestone decomposes, approximately 1.78 kJ of heat is evolved.