Question

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

Answer 1

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

Step-by-step explanation:

The balanced equation for magnesium oxide produced is given by:

2Mg +
`O_(2)` ⇒ 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) =
`(mass given)/(atomic mass of one mole of the substance)`

n=
`(2.490)/(24)`

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

`(2)/(2)` =
`(x)/(0.1037)`

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.