To calculate the enthalpy change during the decomposition of MgCO3, we need to use the balanced equation and the molar masses of the compounds involved.
The balanced equation for the decomposition of MgCO3 is:
MgCO3(s) -> MgO(s) + CO2(g)
First, let's calculate the molar mass of MgCO3:
- Mg = 24.31 g/mol
- C = 12.01 g/mol
- O = 16.00 g/mol
Molar mass of MgCO3 = 24.31 + 12.01 + (3 * 16.00) = 84.31 g/mol
Next, let's calculate the number of moles of MgCO3:
Number of moles = Mass / Molar mass
Number of moles = 54.7 g / 84.31 g/mol = 0.649 mol
Now, we need to use the stoichiometric coefficients from the balanced equation to determine the molar ratio for the enthalpy change:
From the balanced equation: 1 mol of MgCO3 -> 1 mol of MgO
So, the number of moles of MgO formed will be equal to 0.649 mol.
Next, we need to calculate the enthalpy change using the enthalpy of formation values.
The enthalpy change for the decomposition of 1 mol of MgCO3 is given as -1171 kJ/mol.
Enthalpy change = -1171 kJ/mol * 0.649 mol = -759.679 kJ (rounded to three decimal places)
Therefore, the enthalpy change during the decomposition of 54.7g of MgCO3 is approximately -759.679 kJ.
Since the actual answer you mentioned is 76.1 kJ, please double-check the enthalpy change value given in the question or provide any additional information if available.