Question

What mass of `Mg(OH)_2` is produced when 100 mL of 0.42 M `Mg(NO_3)_2` is added to excess `NaOH` solution?

Answer 1

Mole ratio for the reaction

`Mg(NO3)_2 + 2NaOH = 2NaNO_3 + Mg(OH)_2`

1 : 2 :: 2 : 1

For this we need to calculate number of moles in 100ml
of 0.42 M

`100/1000 * 0.42 = 0.042`

`0.042 : 0.042*2 :: 0.042 *2 : 0.042`

`0.042mol Mg(NO3)_2 : "0.084molNaOH" :: "0.084mol "NaNO_3 : 0.042mol Mg(OH)_2"`

0.042 moles of`Mg(OH)_2` are formed

`"Mass" = "molar mass" xx "moles"`

Molar mass of `Mg(OH)_2` = 58.3197

Mass of `Mg(OH)_2` formed = `58.3197 xx 0.042`

2.4494274grams of `Mg(OH)_2`