Question

Magnesium hydroxide, $Mg(OH{)}_2$ is the white milky substance in milk of magnesia. What mass of $Mg(OH{)}_2$ is formed when 15 mL of 0.18 M NaOH are combined with 12 mi. of 0.14 M $MgC{l}_2$ ? The molar mass of $Mg(OH{)}_2$ is 58.3 g $mo{l}^{-1}$.

• A. 0.079 g
• B. 0.097 g
• C. 0.16 g
• D. 0.31 g

Answer 1

The correct option is A 0.079 g

$MgC{l}_2+2NaOH\to Mg(OH{)}_2+2NaCl$

moles of NaOH in 15 ml of 0.18 M solution = $\frac{0.18\ast 15}{1000}$ = 0.0027

moles of$MgC{l}_2$ in 12 ml of 0.14 M solution = 0.00168

according to above equation, for 1 mole of $MgC{l}_2$ we require 1 mole of NaOH

so, NaOH is a limiting reagent

so, moles of $Mg(OH{)}_2$ produced = $\frac{0.0027}{2}$ = 0.0013

so, mass of $Mg(OH{)}_2$ produced = moles * molar mass = 0.0013 * 58.3 = 0.079 g