Chemistry

Chemical Reactions and Equations

684 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 distinguish between black copper oxide and black manganese dioxide practical work chemistry

Write the ionic equation for reaction between manganese dioxide and hydrochloric acid.

  1. $MnO _2(s)+ 2Cl^{-}(aq) \rightarrow Mn^{2+}(aq) + Cl _2(g) + 2H _2O(l)$
  2. $MnO _2{(s)} + 4H^{+}{(aq)} + 2Cl^{-}{(aq)} \rightarrow Mn^{2+}{(aq)} + Cl _2{(g)} + 2H _2O{(l)}$
  3. $MnO _2{(s)} + 3H^{+}{(aq)} + 2Cl^{-}{(aq)} \rightarrow Mn^{2+}{(aq)} + Cl _2{(g)} + 2H _2O{(l)}$
  4. $MnO _2{(s)} + 4H^{+}{(aq)} + 2Cl^{-}{(aq)} \rightarrow Mn^{2+}{(aq)} + Cl{(g)} + 2H _2O({l})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$MnO _2{(s)} + 4H^{+}{(aq)} + 2Cl^{-}{(aq)} \rightarrow Mn^{2+}{(aq)} + Cl _2{(g)} + 2H _2O{(l)}$

Multiple choice reactions of acids and bases properties of acids and bases acids, bases and salts acids and bases chemistry

Which of the following equation represents action of alkali on metal?

  1. $Zn+2NaOH\rightarrow Na _2ZnO _2+H _2$
  2. $Zn+2KOH\rightarrow K _2ZnO _2+H _2$
  3. $Pb+2KOH\rightarrow K _2PbO _2+H _2$

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Alkali reacts with hot concentrated caustic alkalies to give corresponding soluble salt and liberate hydrogen.

Multiple choice reactions of acids and bases properties of acids and bases acids, bases and salts acids and bases chemistry

Which of the following equation is correct for action of alkali on aluminium?

  1. $2Al+2NaOH+2H _2O\rightarrow2NaAlO _2+3H _2$
  2. $2Al+2KOH+2H _2O\rightarrow2KAlO _2+3H _2$
  3. $Al+NaOH+H _2O\rightarrow NaAlO _2+H _2$

  4. Both A and B
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$2Al+2NaOH+2H _2O\rightarrow2NaAlO _2+3H _2$ and $2Al+2KOH+2H _2O\rightarrow2KAlO _2+3H _2$ 

Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

A definite mass of $ H _{2}O _{2} $ is oxidized by excess of acidified $ KMnO _{4} $ and acidified $ K _{2}Cr _{2}O _{7} $, in separate experiments. Which of the following is/are correct statements? 
(K = 39, Cr = 52, Mn = 55 )

  1. Mass of $ K _{2}Cr _{2}O _{7} $ used up will be greater than that of $ KMnO _{4} $
  2. Moles of $ KMnO _{4} $ used up will be greater than that of $ K _{2}Cr _{2}O _{7} $

  3. Equal mass of oxygen gas is evolved in both the experiments.
  4. If equal volumes of both the solutions are used for complete reaction, then the molarities of $ KMnO _{4} $ and $ K _{2}Cr _{2}O _{7} $ solutions are in $6:5$ ratio.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

According to question, reaction of experiment (1) and (2)
(1)$ 5H _{2}O _2+2KMnO _{4}+3H _{2}SO _{4}\rightarrow 5O _{2}+2MnSO _{4}+K _{2}SO _{4}+8H _{2}O $


(2)$ 3H _{2}O _{2}+K _{2}Cr _{2}O _{7}+4H _2SO _{4}\rightarrow Cr _{2}(SO _{4}) _{3}+3O _{2}+K _{2}SO _{4}+7H _{2}O $

(a) According to reaction (1)
5 mole $ H _{2}O _{2} = 2\,mole KMnO _{4} $

1 mole $ H _{2}O _{2} = \dfrac{2}{5} = 0.4\,mole\,KMnO _{4} = 63.2\,g\,KMnO _{4} $

According to reaction (2)
3 mole $ H _{2}O _{2} = 1\,mole\,K _{2}Cr _{2}O _{7} $

1 mole $ H _{2}O _{2} = \dfrac{1}{3} mole\,K _{2}Cr _{2}O _{7} = 98.1 g \, K _{2}Cr _{2}O _{7} $

Mass of $ K _{2}Cr _{2}O _{7}> KMnO _{4} $

(b) Moles of $ KMnO _{4}> $ moles of $ K _{2}Cr _{2}O _{7} (0.333) $

(c)In reaction 1, 5 Mole $ H _{2}O _{2} $ released = 5 mole $ O _{2}\Rightarrow 32\times 5=160g $
In reaction 2, 3 mole $ H _{2}O _{2} $ released = 3 mole $ O _{2}=3\times 32= 96 g$ 

(d) 1 mole of $ H _{2}O _{2} = \dfrac{2}{5} $ moles $ KMnO _{4} $ exp...(1)
1 mole of $ H _{2}O _{2} = \dfrac{1}{3}$ mole $ K _{2}Cr _{2}O _{7} $ exp...(2)
$ \dfrac{KMnO _{4}}{K _{2}Cr _{2}O _{7}} = \dfrac{\dfrac{2}{5}}{\dfrac{1}{3}} = \dfrac{2}{5}\times \dfrac{3}{1} = \dfrac{6}{5}\Rightarrow 6:5 $

Options A, B and D are correct.

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). (B) reacts with organic compounds containing -OH group converting them into chloro derivatives. (B) on hydrolysis gives (C) and is finally converted to phosphoric acid. (A), (B) and (C) are?

  1. $P _{4}$,$PCl _{3}$,$H _{3}PO _{4}$
  2. $P _{4}$,$PCl _{5}$,$H _{3}PO _{3}$
  3. $P _{4}$,$PCl _{5}$,$POCl _{3}$
  4. $P _{4}$,$PCl _{3}$,$POCl _{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A translucent white waxy solid (A) reacts with excess of chlorine to give a yellowish white powder (B). Here B is $PCl _5$.$PCl _5$ reacts with organic compounds containing -OH group converting them into chloro derivative $POCl _3. $.$PCl _5$ on hydrolysis gives (C).C is $POCl _3$ and is finally $POCl _3$ converted to phosphoric acid.

So  (A), (B) and (C) are $P _{4}$,$PCl _{5}$,$POCl _{3}$.

Hence option C is correct.

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

$\mathrm{P}\mathrm{Cl} _{3}$ is prepared by the action of $\mathrm{Cl} _{2}$ on :

  1. $ \mathrm{P} _{2}\mathrm{O} _{3}$
  2. $ \mathrm{P} _{2}\mathrm{O} _{5}$
  3. White $\mathrm{P}$
  4. $\mathrm{H} _{3}\mathrm{P}\mathrm{O} _{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

White P is reactive due to strain.
$P _{4}+6Cl _{2}\ \to \ 4PCl _{3}$
(White P)

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

What is produced as an impurity during the production of ${PCl} _{3}$:

  1. ${PCl} _{5}$
  2. ${POCl} _{3}$
  3. ${PSCl} _{3}$
  4. ${Cl} _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

During the indusrial preparation of ${PCl} _{3}$, ${PCl} _{5}$ or Phosphorus Pentachloride is produced as an impurity. The balanced equation can be written as-
${PCl} _{3}$ ${+}$ ${Cl} _{2}$ ${=}$ ${PCl} _{5}$
Thus this impurity formation can be stopped by continual removal of ${PCl} _{3}$ as it is formed.

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

The products of the hydrolysis of $\mathrm{P}\mathrm{Cl} _{3}$ are :

  1. $\mathrm{H}\mathrm{Cl},\ \mathrm{H} _{3}\mathrm{P}\mathrm{O} _{4}$
  2. $\mathrm{H}\mathrm{Cl},\ \mathrm{H} _{3}\mathrm{P}\mathrm{O} _{3}$
  3. $\mathrm{Cl} _{2},\ \mathrm{H} _{3}\mathrm{P}\mathrm{O} _{3}$
  4. $\mathrm{H}\mathrm{Cl},\ \mathrm{P}\mathrm{O}\mathrm{Cl} _{3},\ \mathrm{H} _{3}\mathrm{P}\mathrm{O} _{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Hydrolysis of $PCl _3$ gives:

$2PCl _3+6H _2O\rightarrow 2H _3PO _3+6H+$

$PCl _3+3H _2O\rightarrow H _3PO _3+3H+$

Option B is correct.

Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

The products formed by the complete hydrolysis of $P{Cl} _{3}$ are:

  1. ${H} _{3}{PO} _{3}$ and $HCl$
  2. $PO{Cl} _{3}$ and $HCl$
  3. ${H} _{3}{PO} _{4}$ and $HCl$
  4. ${H} _{4}{P} _{2}{O} _{7}$ and $HCl$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The products formed by the complete hydrolysis of $PCl _3$ are as follow:
$PC{ l } _{ 3 }+3{ H } _{ 2 }O\rightarrow { H } _{ 3 }P{ O } _{ 3 }+3HCl$

${ H } _{ 3 }P{ O } _{ 3 }$ and $HCl$ are the products formed.
Multiple choice chemistry p- block elements-ii compounds of phosphorus - pcl3 phosphorus halides compounds of phosphorus compounds of phosphorus- pcl5

On hydrolysis, $PCI _3$ gives :

  1. $H _3PO _3$
  2. $H _3PO _4$
  3. $POCI _3$
  4. $HPO _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$PCl _{3}$ on hydrolysis gives $HCl$ and $ H _{3}PO _{3}$

$PCl _{3}+ 3H _{2}O\rightarrow H _{3}PO _{3}+3HCl$

Phosphorus trichloride reacts violently with water forming phosphorous acid.

Option (A) is correct .

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

For the reaction, $2Fe(NO _3) _3+3Na _2CO _3\rightarrow Fe _2(CO _3) _3+6NaNO _3$ initially 2.5 mole of $Fe(NO _3) _2$ and 3.6 mole of $Na _2CO _3$ are taken. If 6.3 mole of $NaNO _3$ is obtained then % yield of given reaction is:

  1. 50

  2. 84

  3. 87.5

  4. 100
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\quad \quad \quad \quad 2Fe{ (N{ O } _{ 3 }) } _{ 3 }+3{ Na } _{ 2 }{ CO } _{ 3 }\longrightarrow { Fe } _{ 2 }{ (C{ O } _{ 3 }) } _{ 3 }+6{ Na }{ NO } _{ 3 }\ initial\quad \quad 2.5\quad \quad \quad \quad \quad 3.6\quad \quad \quad \quad \quad  \quad - \quad \quad \quad \quad -\ after\quad 2.5-2.4=0.1\quad \quad 0\quad \quad \quad \quad \quad \quad 1.2 \quad \quad \quad \quad 6.3\ reaction$

As 2 moles of $Fe(NO _{3}) _{3}$ reacts with 3 moles of $Na _{2}CO _{3}$. 
Thus 2.5 moles of $Fe(NO _{3}) _{3}$ reacts with=$\cfrac{3}{2} \times 2.5$ moles of $Na _{2}CO _{3}$=3.75 moles of $Na _{2}CO _{3}$
As $Na _{2}CO _{3}$ present is 3.6 moles only. Thus, $Na _{2}CO _{3}$ is limiting reagent.
Now, 3 moles of $Na _{2}CO _{3}$ gives=6 moles of $NaNO _{3}$ 
3.6 moles of $Na _{2}CO _{3}$ gives=$\cfrac{6}{3} \times 3.6$ moles of $NaNO _{3}$
3.6 moles of $Na _{2}CO _{3}$ gives=7.2 moles of $NaNO _{3}$
But $NaNO _{3}$ obtained is 6.3 moles.
Thus % yield=$\cfrac{6.3}{7.2} \times 100=87.5 \%$