What mass of $M g (O H)_{2}$ is required to neutralize 25 mL of 0.136 M hydrochloric acid solution?
Molar mass of $M g (O H)_{2} = 58.33 g m o l^{- 1}$

0.248 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.496 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.992 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.98 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.496 g
$M g (O H)_{2} + 2 H C l \to M g C l_{2} + 2 H_{2} O$
Number of moles of $H C l$ $= \frac{M V}{1000} = \frac{0.136 \times 125}{1000} = 0.017$
$∵ 2$ mole $H C I$ is neutralized by $58.33$ g of $M g (O H)_{2}$
$∴ 0.017$ mole $H C l$ will be neutralized by $\frac{58.33}{2} \times 0.017 g$ of $2 M g (O H)_{2} = 0.496 g$

Suggest Corrections

Similar questions

What mass of $M g (O H)_{2}$ is required to neutralise 125 ml of 0.136 M hydrochloric acid solution ?Molar mass of $M g (O H)_{2}$ = 58.33 g $m o l^{- 1}$

Q. What mass of magnesium hydroxide is required to neutralize

125 m L

0.136 M H C l

solution?

(Molar mass of

M g (O H)_{2} = 58.33 g m o l^{- 1}

)

M g (O H)_{2}

in the form of milk of magnesia is used to neutralize excess stomach acid. How many moles of stomach acid can be neutralized by 1 g of

M g (O H)_{2}

?
(Molar mass of

M g (O H)_{2} = 58.33

)

Q. Mass of

H_{2} C_{2} O_{4}

required to reduce 100 ml of 0.008M

K M n O_{4}

IN acidic medium is x g and to neutralize 100 ml of 0.01M

M g {(O H)}_{2}

IS y g, then the relation between x and y is

Q. Magnesium hydroxide,

M g (O H)_{2}

is the white milky substance in milk of magnesia. What mass of

M g (O H)_{2}

is formed when 15 mL of 0.18 M NaOH are combined with 12 mi. of 0.14 M

M g C l_{2}

? The molar mass of

M g (O H)_{2}

is 58.3 g

m o l^{- 1}

Join BYJU'S Learning Program

What mass of Mg(OH)2 is required to neutralize 25 mL of 0.136 M hydrochloric acid solution? Molar mass of Mg(OH)2=58.33gmol−1

The correct option is C 0.496 g Mg(OH)2+2HCl→MgCl2+2H2ONumber of moles of HCl=MV1000=0.136×1251000=0.017∵2 mole HCI is neutralized by 58.33g of Mg(OH)2 ∴0.017 mole HCl will be neutralized by 58.332×0.017g of 2Mg(OH)2=0.496g

What mass of $M g (O H)_{2}$ is required to neutralize 25 mL of 0.136 M hydrochloric acid solution?
Molar mass of $M g (O H)_{2} = 58.33 g m o l^{- 1}$