Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Catalytic Hydrogenation

Question

$C H_{4}$ can be prepared by the reaction of $H_{2} O$ with:
$M g_{2} C_{3}$
$C a C_{2}$
$A l_{4} C_{3}$
$B e_{2} C$

A

A l_{4} C_{3}

B

M g_{2} C_{3}

C

C a C_{2}

D

B e_{2} C

Open in App

Solution

Verified by Toppr

Carbides of Be and Al contain methanide ion $(C^{4 -}$ and thus give $C H_{4}$ gas (due to the diagonal relationship of Be with Al).
$B e_{2} C \to 2 C e^{2 +} + C^{4 -}, A l_{4} C_{3} \to 4 A l^{3 +} + C^{4 -}$
$B e_{2} C + 4 H_{2} O \to C H_{4} + 3 B e (O H)_{2}$
$A l_{4} C_{3} + 12 H_{2} O \to 3 C H_{4} 4 A l (O H)_{3}$
Whereas the carbides of Mg, i.e. $M g_{2} C_{3},$ contain propynide ion $(C_{3}^{4 -}$ ) and hence give propyne.

Was this answer helpful?

0

Similar Questions

Q1

Which is not obtained when metal carbides react with H₂O?

View Solution

Q2

How many of the following carbides on reaction with water will produce a product capable of forming silver mirror, on treatment with Tollen's reagent?

B e_{2} C, C a C_{2}, M g_{2} C_{3}, A l_{4} C_{3}, C H_{4},

View Solution

Q3

One mole each of

C a C_{2}, A l_{4} C_{3} a n d M g_{2} C_{3}

react with

H_{2} O

in separate open flasks at

25^{o} C .

Numerical value of the work done by the system is in the order:
Given reactions:

C a C_{2} (s) + 2 H_{2} O (l) \to C a (O H)_{2} (s) + C_{2} H_{2} (g)

A l_{4} C_{3} (s) + 12 H_{2} O (l) \to 4 A l (O H)_{3} (s) + 3 C H_{4} (g)

M g_{2} C_{3} (s) + 4 H_{2} O (l) \to 2 M g (O H)_{2} (s) + C H_{3} - C \equiv C H (g)

View Solution

Q4

C H_{4}

can be prepared by the reaction of

H_{2} O

with:

View Solution

Q5

S i C, C a C_{2}, B e_{2} C, A l_{2} C_{6}, A l_{4} C_{3}, B_{4} C

How many of the given carbides are methanides?

View Solution