Prepare hydrogen from dil. $H_{2} S O_{4}$ .

2 Z n + H_{2} {S O}_{4} (d i l) \to 2 {Z n S O}_{4} + H_{2}

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Z n + H_{2} {S O}_{4} (d i l) \to {Z n S O}_{4} + H_{2}

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Z n + 2 H_{2} {S O}_{4} (d i l) \to 2 {Z n S O}_{4} + H_{2}

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2 Z n + 2 H_{2} {S O}_{4} (d i l) \to 2 {Z n S O}_{4} + 2 H_{2}

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Solution

The correct option is B $Z n + H_{2} {S O}_{4} (d i l) \to {Z n S O}_{4} + H_{2}$
$Z n + H_{2} {S O}_{4} (d i l) \to {Z n S O}_{4} + H_{2}$

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Similar questions

P b {(N O_{x})}_{2} + H_{2} {S O}_{4} (d i l) \to {P b S O}_{4} ↓ + {y H N O}_{3}

Find x and y.

Q. Balanced equation for the reaction of conc. sulphuric acid with Sucrose is

2 C_{12} H_{22} O_{11} + H_{2} {S O}_{4} (c o n c) \to 12 C + 5 H_{2} O + H_{2} {S O}_{4} (d i l)

1.	Column-A Types of chemical reaction		Column-B Chemical equation
(A)	Combination reaction	(i)	$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
(B)	Oxidation & reduction reaction	(ii)	$2 H_{2} O E l e c t r i c i t y ⟶ 2 H_{2} + O_{2}$
(C)	Decomposition reaction	(iii)	$C a O + C O_{2} \to C a C O_{3}$
(D)	Displacement reaction	(iv)	$H_{2} + C l_{2} \to 2 H C l$
(E)	Double Displacement reaction	(v)	$B a C l_{2} + N a_{2} S O_{4} \to B a S O_{4} ↓ + N a C l$

Q. If

f^{'} (3) = 2, t h e n lim h \to 0 \frac{f (3 + h^{2}) - f (3 - h^{2})}{2 h^{2}}

Given the standard reduction potentials:

Zn²⁺/Zn = - 0.74 V, Cl₂/Cl^- = 1.36 V, H⁺/1/2H₂ = 0 V and Fe³⁺/Fe²⁺ = 0.77 V.
The order of increasing strength as reducing agent is

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Prepare hydrogen from dil.H2SO4.

The correct option is B Zn+H2SO4(dil)→ZnSO4+H2Zn+H2SO4(dil)→ZnSO4+H2

Prepare hydrogen from dil. $H_{2} S O_{4}$ .

The correct option is B $Z n + H_{2} {S O}_{4} (d i l) \to {Z n S O}_{4} + H_{2}$
$Z n + H_{2} {S O}_{4} (d i l) \to {Z n S O}_{4} + H_{2}$