Answer: 15.56 L of oxygen gas reacts to produce 56.1 grams of magnesium oxide at STP.
Step-by-step explanation:
The given chemical equation represents the reaction between magnesium (Mg) and oxygen (O2) to form magnesium oxide (MgO) with a stoichiometric ratio of 2:1 between Mg and O2. This means that for every 2 moles of Mg that reacts, 1 mole of O2 is consumed.
The molar mass of MgO is 40.3 g/mol (24.3 g/mol for Mg + 16.0 g/mol for O). Therefore, the number of moles of MgO produced can be calculated as follows:
Number of moles of MgO = Mass of MgO / Molar mass of MgO
Number of moles of MgO = 56.1 g / 40.3 g/mol
Number of moles of MgO = 1.39 mol
Since the stoichiometric ratio of Mg to O2 is 2:1, we can calculate the number of moles of O2 consumed as follows:
Number of moles of O2 = (Number of moles of MgO) / 2
Number of moles of O2 = 1.39 mol / 2
Number of moles of O2 = 0.695 mol
At STP (standard temperature and pressure), one mole of any ideal gas occupies 22.4 L. Therefore, the volume of O2 consumed can be calculated as follows:
Volume of O2 consumed = Number of moles of O2 x 22.4 L/mol
Volume of O2 consumed = 0.695 mol x 22.4 L/mol
Volume of O2 consumed = 15.56 L
Therefore, 15.56 L of oxygen gas reacts to produce 56.1 grams of magnesium oxide at STP.