How many grams of magnesium oxide can be produced if 2.490 Mg react with excess oxygen?

2.488 grams of MgO is produced when 2.49 grams of Mg reacts with excess of oxygen.

Explanation:

The balanced equation for magnesium oxide produced is given by:

2Mg + ⇒ 2MgO

it can be seen from the equation that 2 moles of Mg will produce 2 moles of MgO

from the mass of the MgO given moles of Mg can be calculated, the atomic  mass of Mg is 24 gm/mole

the formula to calculate number of moles:

number of moles (n) =

                               n=

                                n = 0.1037 moles of Mg reacts with excess of oxygen.

Applying stoichiometry,

2 moles of Mg yields 2 mole of MgO

0.1037 moles of Mg yields x mole of MgO

=  

2x = 2 × 0.1037

x = 0.1037 moles of MgO is formed.

The mass of MgO produced could be known by multiplying atomic weight with number of moles.

mass = 0.1037 × 24

         = 2.488 grams of MgO wiil be produced.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.