How many grams of magnesium oxide are formed when 15.4 L of oxygen gas, measured at STP, completely reacts with magnesium metal according to this reaction?

2Mg(s)+O2(g)→2MgO(s)

55 g OF MgO WILL BE FORMED IF 15.4 L OF OXYGEN WERE TO BE USED AT STP.

Explanation:

Equation for the reaction;

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

From the balanced equation, 2 moles of Mg reacts with 1 mole of Oxygen to yield 2 moles of MgO

1 mole of O2 yields 2 moles of MgO

At STP, 1 mole of a gas = 22.4 dm3

So therefore, 22.4 dm3 of Oxygen formed 2 moles of MgO

Calculate the molar mass of Mg0

(Mg = 24, O = 16)

RMM = (24 + 16) g/mol

RMM = 40 g/mol

Since 2 moles of MgO is formed, to calculate the mass of MgO formed, we multiply the molar mass by the number of moles.

Mass of MgO formed = 40 g/mol * 2 moles

Mass = 80 g

So therefore,

22.4 dm3 of oxygen formed 80 g pf MgO

If 15.4 L of oxygen were to be used, how many grams of MgO will be formed;

22.4 L = 80 g of MgO

15.4 L of O2 = x

x = 80 * 15.4 / 22.4

x = 1232 / 22.4

x =55 g

55g of MgO will be formed at STP if 15.4 L of oxygen gas were to be used.

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