Chemistry

Chemical Reactions and Equations

674 Questions

Chemical reactions and equations form a foundational chemistry topic where students identify reaction products, balance chemical formulas, and classify reaction types. It covers critical mechanisms like Markovnikov's rule, precipitation, and endothermic or exothermic processes. These questions are highly common in general science sections of state and central government competitive exams.

Endothermic reactionsPrecipitation reactionsDehydrohalogenation of alkyl halidesChemical equation balancingReaction product identification

Chemical Reactions and Equations Questions

Multiple choice chemistry the p-block elements - group 13 study of orthoboric acid some important compounds of boron study of boron

$H _{3}BO _{3}\overset{T _{1}}{\rightarrow}X\overset{T _{2}}{\rightarrow}Y\overset{red:hot}{\rightarrow}B _{2}O _{2}$


 if $T _{1}< T _{2}$ then X and Y respectively are-

  1. $X=$ Metaboric acid and $Y=$ Tetraboric acid
  2. $X=$ Tetraboric acid and $Y=$ Metaboric acid
  3. $X=$ Borax and $Y=$ Metaboric acid
  4. $X=$ Tetraboric acid and $Y=$ Borax
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Effect of temperature at $100^{\circ}C \,,\,H _{3}BO _{3}$ losses water and convert into metaboric acid.

$H _{3}BO _{3}\xrightarrow{100^{\circ}C}HBO _{2}+H _{2}O$

metaboric acid form tetraboric acid on heating at $160^{\circ}C$

$4HBO _{2}\xrightarrow{160^{\circ}C}H _{2}B _{4}O _{7}+H _{2}O$

On strong heating, $B _{2}O _{3}$ is produced

$H _{2}B _{4}O _{7}\rightarrow 2B _{2}O _{3}+H _{2}O$

Hence, option $A$ is correct.
Multiple choice chemistry the p-block elements - group 13 study of orthoboric acid some important compounds of boron study of boron

Which of the following compounds are formed when $BCl _3$ is treated with water?

  1. $H _3BO _3$
  2. $B _2H _6$
  3. $B _2O _3$
  4. $HBO _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$B{Cl} _3$ hydrolyzes readily to give hydrochloric acid and boric acid.

$B{Cl} _3 + 3H _2O \; \xrightarrow{hydrolyzes} \; B{(OH)} _3 + 3HCl$

Hence, the correct answer is option $\text{A}$.
Multiple choice chemistry the p-block elements - group 13 study of orthoboric acid some important compounds of boron study of boron

The reaction,

$B(OH) _3+NaOH\rightarrow Na[B(OH) _4]$ can be made to proceed in forward direction by :

  1. adding cis-1, 2-diol.

  2. adding borax.

  3. adding trans-1, 2-diol.
  4. adding $Na _2HPO _4$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If polyhydroxy compound like glycerol, mannitol or sugar are added to titration mixture than it can be titrated with $NaOH$.

Due to formation of chelated complex, the reaction moves in forward direction.

$B(OH) _{3}+NaOH\rightarrow Na[B(OH) _{4}]$
Multiple choice chemistry the p-block elements - group 13 study of orthoboric acid some important compounds of boron study of boron

$B _{2}Ca _{6}O _{11}+ Na _{2}CO _{3}\overset{\Delta }{\rightarrow}\left [ X \right ]+CaCO _{3}+NaBO _{2}$ (unbalanced equation)
Correct choice(s) for [X] is/are:

  1. structure of anion of crystalline [X] has one boron atom sp$^{3}$ hybridised and other three boron atoms sp$^{2}$
    hybridised

  2. X with NaOH (aq.) gives a compound which on reaction with hydrogen peroxide in alkaline medium yields
    a compound used as brightner in soaps
  3. hydrolysis of [X] with HCl or H$ _{2}$SO$ _{4}$ yields a compound which on reaction with HF gives fluoroboric acid

  4. [X] on heating with chromium salts in oxidising flame gives green coloured bead in cold
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

Compound [X] is Na$ _{2}$B$ _{4}$O$ _{7}$
(A) two boron atoms are $sp^{2}$ while other two boron atoms are $sp^{3}$ hybridised.
(B) product is sodium peroxoborate which is used as brightner in soaps.
(C) $H _{3}BO _{3}$ is formed which with HF gives $HBF _{4}$.
(D) Green bead in oxidising as well as in reducing flame in cold.

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

Identify the process to which the following reaction belongs : 


$\displaystyle { Al } _{ 2 }{ O } _{ 3 }.{ 2H } _{ 2 }O+{ Na } _{ 2 }{ CO } _{ 3 }\longrightarrow { 2NaAlO } _{ 2 }+{ 2H } _{ 2 }O+{ CO } _{ 2 }$

$\displaystyle { 2NaAlO } _{ 2 }+{ 2H } _{ 2 }O+{ CO } _{ 2 } \underrightarrow { 50-{ 60 }^{ o }C } { Al } _{ 2 }{ O } _{ 3 }.{ 2H } _{ 2 }O+{ Na } _{ 2 }{ CO } _{ 3 }$

  1. Hall's process

  2. Baeyer's process

  3. Serpeck's process

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hall's Process: (Red bauxite)

Bauxite one + $Na _2CO _3 \xrightarrow{Fused} NaAlO _2$
                                                          $\downarrow $ extracted with water
                                                    Solution
                                                          $\downarrow$ warmed $50^o - 60^oC$
                                                   $CO _2$ is circulated
                                                          $\downarrow$
                                                   $Al(OH) _3 + Na _2CO _3$

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

The metal that cannot displace hydrogen from dilute hydrochlocric acid is 

  1. aluminium

  2. iron

  3. copper

  4. zinc

  5. magnesium
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Metals below hydrogen in the reactivity series, such as copper, cannot displace hydrogen from dilute acids. Aluminium, iron, zinc, and magnesium are all more reactive than hydrogen.

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

Which of the following is produced by the reaction of zinc metal, $Zn$ and hydrochloric acid, $HCl$?
I. $H _2(g)$
II. $Cl _2(g)$
III. $Zn^{2+}(aq)$

  1. II only

  2. III only

  3. I and II only

  4. I and III only

  5. I, II, and III
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle H _2$ and $\displaystyle Zn^{2+}$ is produced by the reaction of zinc metal, $Zn$, and hydrochloric acid, $HCl$.
$\displaystyle Zn + 2HCl  \rightarrow ZnCl _2 + H _2$

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements

The addition of dilute $HCl$ to unknown metal produced a transparent gas. What is the likely identity of this gas?

  1. $Cl _{2}$
  2. $H _{2}$
  3. $O _{2}$
  4. $CO _{2}$
  5. $NO _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Metals that come before hydrogen in the reactivity series form their corresponding chlorides when heated with HCl and liberate hydrogen gas.

Eg:
$Ca+2HCl\rightarrow { CaCl } _{ 2 }+{ H } _{ 2 }$

Multiple choice chemistry p- block elements-ii hydrogen chloride chlorine - 17 group p-block elements


 Appearance  Reaction with dilute $HCl$  Reaction with dilute $HNO _3$
 Unknown metal #1  Dull gray solid with white oxide coating Dissolved with bubbles of clear gas   Dissolved with bubbles of clear gas
 Unknown metal #2  Solid; lustrous, smooth silver-gray surface  No reaction  Dissolved with bubbles of orange gas

Identify the orange gas produced on the addition of dilute $HN{O} _{3}$ to unknown metal #2.

  1. ${Cl} _{2}$
  2. ${H} _{2}$
  3. ${O} _{2}$
  4. $C{O} _{2}$
  5. $N{O} _{2}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Silver has a lustrous smooth gray surface.

It reacts with $HNO _3$ as;
$3Ag+2{ HNO } _{ 3 }\rightarrow { 4AgNO } _{ 3 }+{ NO } _{ 2 }+{ H } _{ 2 }O$.
Thus the orange gas produced is $NO _2$.

Multiple choice chemistry substances in common use preparation, properties and uses of baking soda chemical from common salt compounds of carbon

${NaHCO _3+?\rightarrow NaCl+CO _2+H _2O}$ :

  1. ${H NO _3}$
  2. $HCl$
  3. ${Cl _2}$
  4. ${H _2SO _4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When sodium bicarbonate reacts with $HCl$, chlorine binds with sodium and forms sodium chloride and bicarbonate ion binds with hydrogen ion of $HCl$ and forms carbonic acid. The carbonic acid further dissociates to water and carbon dioxide.