how many liters of oxygen are necessary for the combustion of 124 g of magnesium at stp?

Answer 1

Since the conditions stated are STP (standard temperature and pressure), we can use the fact that 1 mole of gas (ideal) will equal 22.4 liters. But first, we must write a correctly balanced equation for the reaction taking place:

2Mg(s) + O₂(g) ==> 2MgO ... balanced equation

Find moles of Mg present:

124 g Mg x 1 mole Mg/24.31 g = 5.100 moles Mg

Find moles O₂ needed to react with 5.100 moles of Mg:

5.100 moles Mg x 1 mole O₂ / 2 mole Mg = 2.55 moles O₂

Find volume of O₂ represented by 2.55 moles of O₂ at STP:

2.55 moles O₂ x 22.4 liters / mole = 57.1 liters of O₂