Magnesium hydroxide, `Mg(OH)_(2)` is the white milky substance in milk of magnesia. What mass of `Mg(OH)_(2)` is formed when 15mL of 0.18M of NaOH combined with 12mL of 0.14M `MgCl_(2)`? The molar mass of `Mg(OH)_(2)` is 58.3`g mol^(-1)`

Options

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

Answer

A

Explanation

`n_(MgCl_(2))=(MV)/(1000)=(12xx0.14)/(1000)=1.68xx10^(-3)`
`n_(NaOH)=(15xx0.18)/(1000)=2.7xx10^(-3)`
NaOH will be limiting reagent because on complete consumption NaOH give least amount of `Mg(OH)_(2)`.
`"Mass of "Mg(OH)_(2)=(1)/(2)xx2.7xx10^(-3)xx58.3=0.079g`