MgCO2(s)-> MgO9(s) + CO2(g)
A. what is delta H reaction when 3.35 mol of CO2 react with excess MgO?
B. What is delta H reaction when 305.5 g of MgCO3 decomposes.
dHrxn = (n*dHf products) - (n*dHf reactants)
dHf = delta H formation. Look this up in your text/notes.
To determine the enthalpy change (ΔH) for the given reactions, we need to find the balanced chemical equations and then use the corresponding values of enthalpy change.
A. Reaction: MgCO2(s) → MgO(s) + CO2(g)
Since the reaction gives the coefficients of the reactants and products, we can use it to calculate the ΔH.
The balanced equation is:
2 MgCO2(s) → 2 MgO(s) + 2 CO2(g)
ΔH for the balanced equation is given as 490.2 kJ.
Now, we can calculate the ΔH for 3.35 mol of CO2 using the equation:
ΔH reaction = ΔH / n
Where ΔH is the ΔH for the balanced equation and n is the stoichiometric coefficient of CO2.
n (CO2) = 2, since the stoichiometric coefficient of CO2 is 2.
ΔH reaction = 490.2 kJ / 2 = 245.1 kJ
Therefore, the ΔH reaction when 3.35 mol of CO2 react with excess MgO is 245.1 kJ.
B. Reaction: MgCO3(s) → MgO(s) + CO2(g)
We need to convert the given mass of MgCO3 to moles before calculating the ΔH.
The molar mass of MgCO3 is:
Mg: 24.31 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 atoms)
Molar mass of MgCO3 (24.31 + 12.01 + (3 × 16.00)) = 84.31 g/mol
Now, we can calculate the number of moles of MgCO3:
Moles (MgCO3) = mass / molar mass
= 305.5 g / 84.31 g/mol
= 3.628 mol
Since the stoichiometric coefficient of MgCO3 in the balanced equation is 1, the number of moles of MgCO3 is the same as the coefficient.
ΔH reaction = ΔH
Hence, we need to find the enthalpy change for the given reaction. The given information is not sufficient to determine the enthalpy change.
To determine the value of ΔH (enthalpy change) for the given reactions, you need to use the given balanced chemical equation and the corresponding enthalpy values (ΔH) of the reactants and products involved.
Before calculating the enthalpy change, let's correct the chemical formula for magnesium carbonate (MgCO2) to MgCO3, as per the reaction equation:
MgCO3(s) → MgO(s) + CO2(g)
A. To find ΔH when 3.35 mol of CO2 react with excess MgO:
1. Identify the stoichiometric ratio between CO2 and MgCO3 in the balanced equation. It is 1:1.
2. Determine the molar mass of CO2:
Molar mass (CO2) = [12.01 g/mol (C) + 2(16.00 g/mol) (O)] = 44.01 g/mol
3. Calculate the mass of CO2:
Mass (CO2) = 3.35 mol × 44.01 g/mol = 147.2935 g
4. Since the reaction is exothermic (releases heat), the enthalpy change (ΔH) will be negative.
Therefore, to find ΔH, you can refer to a reliable source or use experimental enthalpy data for the specific reaction.
B. To find ΔH when 305.5 g of MgCO3 decomposes:
1. Determine the molar mass of MgCO3:
Molar mass (MgCO3) = [24.31 g/mol (Mg) + 12.01 g/mol (C) + 3(16.00 g/mol) (O)] = 84.31 g/mol
2. Calculate the number of moles of MgCO3:
Moles (MgCO3) = Mass (MgCO3) / Molar mass (MgCO3)
= 305.5 g / 84.31 g/mol = 3.6264 mol
3. Apply the stoichiometric ratio from the balanced equation:
As per the balanced equation, the stoichiometric ratio between MgCO3 and MgO is 1:1.
4. Since the reaction is exothermic (releases heat), the enthalpy change (ΔH) will be negative.
Again, to determine the actual value of ΔH, you can refer to a reliable source or use experimental enthalpy data specific to the reaction.