In the industrial production of which of the following, molecular hydrogen is obtained as a byproduct?
- NaOH
- NaCl
- Na metal
- \(Na_2CO_3\)
The Correct Option is A
Solution and Explanation
Molecular hydrogen is produced as a byproduct in the industrial production of NaOH by electrolysis of aq \(NaCl\) solution
\(NaCl → Na^+ + Cl^–\)
\(H_2O ⇌ H^+ + OH^-\)
Cathode : \(2H_2O + 2e → H_2 + 2OH^–\)
Anode : \(2Cl^– → Cl_2 +2e\)
\(NaOH\) is crystallised from the remaining part of electrolyte.
Top Questions on preparation of dihydrogen
For the separation of two immiscible liquids, which method is used?
- BITSAT - 2023
- Chemistry
- preparation of dihydrogen
- The production of dihydrogen gas via water-gas shift reaction $ CO(g) + H_{2}O(g) { ->[{\Delta}][{Catalyst}] } CO_{2}(g) +H_{2}(g) $ The $ CO_{2} $ gas is removed by scrubbing with solution of
- AMUEEE - 2014
- Chemistry
- preparation of dihydrogen
- Hydrogen can be prepared by the action of dil $H_{2}SO_{4}$ on
- AIIMS - 2007
- Chemistry
- preparation of dihydrogen
- Which does not replace hydrogen from $ HCl $ ?
- Rajasthan PMT - 2007
- Chemistry
- preparation of dihydrogen
- Which one of the following can produce hydrogen when treated with metallic sodium?
- Rajasthan PMT - 2006
- Chemistry
- preparation of dihydrogen
Questions Asked in JEE Main exam
In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
- JEE Main - 2025
- Electromagnetic Field (EMF)
- The value of \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \) is:
- JEE Main - 2025
- Trigonometric Identities
- Ice at \( -5^\circ C \) is heated to become vapor with temperature of \( 110^\circ C \) at atmospheric pressure. The entropy change associated with this process can be obtained from:
- JEE Main - 2025
- Thermodynamics
- Let C be the circle of minimum area enclosing the ellipse E: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with eccentricity \( \frac{1}{2} \) and foci \( (\pm 2, 0) \). Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 29 is parallel to the major axis and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:
- JEE Main - 2025
- Coordinate Geometry
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to: