How many grams of MgO are produced during an enthalpy change of -199 kJ? 2 Mg(s) + O2(g) → 2 MgO(s) ΔH = -1204 kJ

To calculate the number of grams of MgO produced during the enthalpy change, we need to use the concept of stoichiometry.

First, we need to determine the stoichiometry from the balanced equation:
2 Mg(s) + O2(g) → 2 MgO(s)

This tells us that 2 moles of magnesium (Mg) reacts with 1 mole of oxygen (O2) to produce 2 moles of magnesium oxide (MgO).

Next, we need to use the given enthalpy change, ΔH = -1204 kJ, to calculate the moles of magnesium oxide produced.

Since the given ΔH is for the reaction of 2 moles of magnesium oxide, we can set up the following proportion:

ΔH (kJ) / moles of magnesium oxide = -1204 kJ / 2 moles

Now, we can solve for the moles of magnesium oxide:

moles of magnesium oxide = (-199 kJ / -1204 kJ) * 2 moles

moles of magnesium oxide = 0.3289 moles

Finally, we need to convert the moles of magnesium oxide to grams using the molar mass of MgO.

The molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol

grams of magnesium oxide = moles of magnesium oxide * molar mass of MgO

grams of magnesium oxide = 0.3289 moles * 40.31 g/mol

grams of magnesium oxide = 13.26 grams

Therefore, 13.26 grams of MgO are produced during the enthalpy change of -199 kJ.