2 Mg(s) + O2(g)= 2 MgO(s) ÄH= -1204 kJ

How many grams of MgO are produced during an enthalpy change of 96.0 kJ?

To find the number of grams of MgO produced during an enthalpy change of 96.0 kJ, we need to use the given enthalpy change and the molar mass of MgO.

1. Determine the number of moles of MgO produced using the enthalpy change:
- Given: ÄH = -1204 kJ
- Desired: ÄH = 96.0 kJ
- Equation: ÄH = (n × ÄH°) / balanced coefficient
- Rearranging the equation: n = (ÄH × balanced coefficient) / ÄH°
- Substituting the values: n = (96.0 kJ × 2) / -1204 kJ

2. Calculate the molar mass of MgO:
- Molar mass (MgO) = atomic mass (Mg) + atomic mass (O)
- Look up the atomic masses of Mg and O in the periodic table and add them together.

3. Calculate the mass of MgO produced:
- Mass = number of moles × molar mass

Let's calculate:

1. n = (96.0 kJ × 2) / -1204 kJ
n = -0.317 moles

2. Molar mass (MgO) = atomic mass (Mg) + atomic mass (O)
Molar mass (MgO) = 24.31 g/mol + 16.00 g/mol
Molar mass (MgO) = 40.31 g/mol

3. Mass = number of moles × molar mass
Mass = -0.317 mol × 40.31 g/mol
Mass = -12.78 g

Since mass cannot be negative, we made an error in the calculations. Please double-check the values given or calculations made.

To find the number of grams of MgO produced during an enthalpy change of 96.0 kJ, we need to use stoichiometry and the given enthalpy change (∆H) value.

The balanced equation for the reaction is:

2 Mg(s) + O2(g) -> 2 MgO(s)

From the balanced equation, we can see that 2 moles of MgO are produced for every 2 moles of Mg reacted.
Since the balanced coefficients relate the number of moles of reactants and products, we can use them to convert between moles and grams using the molar mass of the substance.

The molar mass of Mg is approximately 24.31 g/mol, and the molar mass of MgO is approximately 40.31 g/mol.

To solve the problem, we follow these steps:

Step 1: Calculate the number of moles of MgO produced from the given enthalpy change:
∆H = -1204 kJ (given)
∆H for the reaction is -1204 kJ when 2 moles of MgO are produced.

-1204 kJ / 2 mol MgO = -602 kJ/mol MgO

Step 2: Calculate the number of moles of MgO produced for the given enthalpy change (∆H) of 96.0 kJ:
-602 kJ/mol MgO = 96.0 kJ / x mol MgO

Cross multiply and solve for x:

-602 kJ * x = 96.0 kJ * 1
x = (96.0 kJ * 1) / -602 kJ ≈ -0.159 mol MgO

Note: The negative sign indicates the reverse direction of the reaction. In this case, it represents the reaction going from MgO to Mg and O2.

Step 3: Convert moles of MgO to grams by multiplying with the molar mass of MgO:
-0.159 mol MgO * 40.31 g/mol = -6.42 g MgO

Since mass cannot be negative, we can ignore the negative sign and say that approximately 6.42 grams of MgO are produced during an enthalpy change of 96.0 kJ.