# 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?

Feb 19, 2017

Mole ratio for the reaction

$M g {\left(N O 3\right)}_{2} + 2 N a O H = 2 N a N {O}_{3} + M g {\left(O H\right)}_{2}$

1 : 2 :: 2 : 1

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

$\frac{100}{1000} \cdot 0.42 = 0.042$

$0.042 : 0.042 \cdot 2 : : 0.042 \cdot 2 : 0.042$

$0.042 m o l M g {\left(N O 3\right)}_{2} : \text{0.084molNaOH" :: "0.084mol "NaNO_3 : 0.042mol Mg(OH)_2}$

0.042 moles of$M g {\left(O H\right)}_{2}$ are formed

$\text{Mass" = "molar mass" xx "moles}$

Molar mass of $M g {\left(O H\right)}_{2}$ = 58.3197

Mass of $M g {\left(O H\right)}_{2}$ formed = $58.3197 \times 0.042$

2.4494274grams of $M g {\left(O H\right)}_{2}$