The enthalpy change for the reaction can be calculated using the stoichiometry of the reaction and the molar mass of calcium carbonate (CaCO3).

1. First, we need to find the molar mass of CaCO3. The molar mass of Ca is approximately 40.08 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol. Therefore, the molar mass of CaCO3 is 40.08 g/mol + 12.01 g/mol + (3*16.00 g/mol) = 100.09 g/mol.

2. Next, we need to find the number of moles of CaCO3 formed. This can be found by dividing the mass of CaCO3 formed by its molar mass. So, 380 g ÷ 100.09 g/mol = 3.79 moles of CaCO3.

3. The enthalpy change given in the problem is for the formation of 1 mole of CaCO3. Therefore, the enthalpy change for the formation of 3.79 moles of CaCO3 is -69.1 kJ/mol * 3.79 mol = -261.7 kJ.

So, the enthalpy change for the formation of 380 g of CaCO3 is -261.7 kJ.

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The enthalpy change for the reaction can be calculated using the stoichiometry of the reaction and the molar mass of calcium carbonate (CaCO3).

1. First, we need to find the molar mass of CaCO3. The molar mass of Ca is approximately 40.08 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol. Therefore, the molar mass of CaCO3 is 40.08 g/mol + 12.01 g/mol + (3*16.00 g/mol) = 100.09 g/mol.

2. Next, we need to find the number of moles of CaCO3 formed. This can be found by dividing the mass of CaCO3 formed by its molar mass. So, 380 g ÷ 100.09 g/mol = 3.79 moles of CaCO3.

3. The enthalpy change given in the problem is for the formation of 1 mole of CaCO3. Therefore, the enthalpy change for the formation of 3.79 moles of CaCO3 is -69.1 kJ/mol * 3.79 mol = -261.7 kJ.

So, the enthalpy change for the formation of 380 g of CaCO3 is -261.7 kJ.

Knowee AI · Accepted Answer

The enthalpy change for the reaction can be calculated using the stoichiometry of the reaction and the molar mass of calcium carbonate (CaCO3).

1. First, we need to find the molar mass of CaCO3. The molar mass of Ca is approximately 40.08 g/mol, C is 12.01 g/mol, and O is 16.00 g/mol. Therefore, the molar mass of CaCO3 is 40.08 g/mol + 12.01 g/mol + (3*16.00 g/mol) = 100.09 g/mol.

2. Next, we need to find the number of moles of CaCO3 formed. This can be found by dividing the mass of CaCO3 formed by its molar mass. So, 380 g ÷ 100.09 g/mol = 3.79 moles of CaCO3.

3. The enthalpy change given in the problem is for the formation of 1 mole of CaCO3. Therefore, the enthalpy change for the formation of 3.79 moles of CaCO3 is -69.1 kJ/mol * 3.79 mol = -261.7 kJ.

So, the enthalpy change for the formation of 380 g of CaCO3 is -261.7 kJ.

Calcium hydroxide reacts with carbon dioxide to form calcium carbonate. The reaction for this process is:Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O(g), ΔH = -69.1 kJWhat is the enthalpy change if 380 g of calcium carbonate is formed?

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