a) Is this reaction endothermic or exothermic?
The reaction is endothermic. We can determine this by examining the sign of the enthalpy change (ΔH). In this reaction, ΔH = 1204 kJ, which is positive. This indicates that energy is absorbed from the surroundings during the reaction, making it endothermic.
b) How many grams of MgO are produced during an enthalpy change of 96.0 kJ?
To find the grams of MgO produced from an enthalpy change of 96.0 kJ, we first establish the relationship between the enthalpy change and the number of moles of MgO formed. The given ΔH of 1204 kJ corresponds to the formation of 2 moles of MgO. Thus, the energy required to produce 1 mole of MgO is 1204 kJ / 2 = 602 kJ.
Now, we can calculate the number of moles of MgO produced for 96.0 kJ:
Number of moles = 96.0 kJ / 602 kJ/mol ≈ 0.159 moles.
Next, we convert moles of MgO to grams using its molar mass. The molar mass of MgO (Mg = 24.31 g/mol, O = 16.00 g/mol) is approximately 40.31 g/mol.
So, grams of MgO = 0.159 moles × 40.31 g/mol ≈ 6.41 grams.
c) How many kilojoules of heat are absorbed when 7 grams of MgO are produced?
First, we determine how many moles of MgO correspond to 7 grams:
Moles of MgO = 7 g / 40.31 g/mol ≈ 0.173 moles.
Since producing 2 moles of MgO requires 1204 kJ, producing 1 mole of MgO requires 602 kJ. Thus, the heat absorbed for 0.173 moles of MgO is calculated as:
Heat absorbed = 0.173 moles × 602 kJ/mol ≈ 104.31 kJ.
Therefore, when 7 grams of MgO are produced, approximately 104.31 kJ of heat are absorbed.