CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Arrhenius Acid

Find the numb...

Question

Find the number of moles of $N a O H$ required to neutralize 448 g equimolar mixture of $N a_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ .

Open in App

Solution

Given, total mass of $N a_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ be 448 g.
Let, moles of each be x.
So, $x \times 134 + x \times 90 = 448 \Rightarrow x = 2$
Equivalents of $N a O H$ required
= Equivalent of $H_{2} C_{2} O_{4}$ (only for $H_{2} C_{2} O_{4})$
Moles of $N a O H$ $\times$ n-factor = mole of $H_{2} C_{2} O_{4}$ $\times$ n -factor
Moles of $N a O H = 2 \times 2 = 4$
(n-factor of $N a O H$ is 1 and n-factor of $H_{2} C_{2} O_{4}$ is 2)

Suggest Corrections

thumbs-up

1

Similar questions

Q. A mixture of

N a_{2} C_{2} O_{4}

K H C_{2} O_{4} . H_{2} C_{2} O_{4}

required equal volumes of

0.2 M

K M n O_{4}

0.2 M

N a O H

separately for complete titration. The mole ratio of

N a_{2} C_{2} O_{4}

K H C_{2} O_{4} . H_{2} C_{2} O_{4}

in the mixture is:

2

mole, equimolar mixture of

N a_{2} C_{2} O_{4}

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

V_{1} L

0.1 M K M n O_{4}

in acidic medium for complete oxidation. The same amount of the mixture required

V_{2} L

0.2 M N a O H

for neutralization. The ratio of

V_{1}

V_{2}

B a F_{2}

is treated with two moles of

H_{2} S O_{4}

.To make the resulting mixture neutral,

N a O H

is added. Moles of

N a O H

required in this process is :

Q. How many moles of

N a O H

are required to neutralize

1.2

H_{2} S O_{4}

?

Q. A mixture containing 2 moles of

K_{2} C_{2} O_{4}

K H C_{2} O_{4}

titrated separately against

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

and NaOH. The moles of acid and base required respectively to reach endpoint would be:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Acids and Bases

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon