Chemistry

Mole Concept and Stoichiometry

354 Questions

The mole concept is a core area of chemistry focusing on the quantitative relationships between reactants and products in chemical reactions. Questions cover molecular mass calculations, stoichiometric conversions, and determining the number of molecules. This topic is essential for most science entrance and competitive examinations.

molecular mass calculationsstoichiometric conversionsmole fraction problemsheavy water reactionsoxidation calculations

Mole Concept and Stoichiometry Questions

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

Total number of electrons present in $14\ gm$ of nitrogen gas is:

$[N _A = 6\times 10^{23}]$

  1. $6\times { 10 }^{ 23 }$
  2. $4.2\times { 10 }^{ 24 }$
  3. $3\times { 10 }^{ 23 }$
  4. $4.2\times { 10 }^{ 23 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

14g of N2 gas is 0.5 moles (molar mass 28g/mol). Each N2 molecule has 14 electrons (7 per nitrogen atom), so 0.5 moles of N2 contains 0.5 * 6 * 10^23 * 14 = 4.2 * 10^24 electrons.

Multiple choice chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

What is the total number of electrons present in 1.6 g of methane?

  1. $6.023 \times 10^{23}$

  2. 16
  3. $12.04 \times 10^{23}$
  4. $6.023 \times 10^{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
1 mole methane = 16 g
Electrons in 1 mole of methane are $=6+4 \times 1=10$
Mole of methane in 1.6 g=$\cfrac{mass}{molar\ mass}$
$=\cfrac{1.6}{16}=0.1$ mole
$\therefore$ 0.1 mol methane contains $0.1 \times 10$mol electron=1 mole of electron
1 mole electron=$6.023 \times 10^{23}$ electrons
Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

The mass of carbon anode consumed (giving of carbon dioxide) in the production of $270 \ kg$ aluminium metal bum bauxite by Hall process is: (Atomic mass $Al = 27$)

  1. $180$ kg
  2. $270$ kg
  3. $540$ kg
  4. $90$ kg
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The chemical reaction for the Hall process is given by

$2Al _2O _3+3C\rightarrow 4Al+3CO _2$
Moles of aluminium produced$=\dfrac{270000}{27}=10000$
According to the reaction,
$10000 \; kg$ moles of aluminium can be produced by consuming $2500\times3=7500 \ moles $ of Carbon
Hence mass of carbon=$12\times 7500=90000\ g=90\ kg$

Therefore, D is the correct answer

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

When sample of baking soda is strongly ignited in crucible it suffered loss in weight of  $3.1 g $  the mass of baking soda is- 

  1. $ 16.8 g $
  2. $ 8.4 g $
  3. $ 11.6 g $
  4. $4.2 g $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The reaction is:
$2NaHCO _3 \rightarrow Na _2CO _3+H _2O+CO _2$

The loss in weight is due to water and carbon dioxide.

$2$ moles $(2\times 84 g)$ of sodium bicarbonate gives $ 1$  mole $( 18  g)$ of water and $ 1$ mole of carbon dioxide $( 44 g)$.

Thus, decomposition of $ 168   g$ of sodium bicarbonate will result in loss in weight of $18+44=62g$

The loss in weight of $3.1$ g corresponds to $\dfrac{168}{62}\times 3.1=8.4g$ of sodium bicarbonate. 

Option B is correct.
Multiple choice physics physical quantities and measurement measurement of small and large distances measurement of distance unit of fundamental quantities

100 kg of Silver has volume $10{ m }^{ 3 }.$ The density of silver in C.G.S is

  1. $10{ g/cm }^{ 3 }$
  2. $0.05{ g/cm }^{ 3 }$
  3. $0.01{ g/cm }^{ 3 }$

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

The mass of silver $= 100 kg$

Volume of the cube $= 10 m^3$

We know that the density of a substance is given by the following formula

$Density = \dfrac{Mass}{Volume}$

Density of the silver $= \dfrac{100}{10} kg/m^3=10kg/m^3$

Now we know that $1 kg = 1000 gm = 10^3 gm$

and $1 m = 100 cm$

$⇒ 1 m^3 = 100^3 \,cm^3 = 10^6 cm^3$

Therefore, the density of silver in cgs $= \dfrac{10\times 10^3}{10^6} gm/cm^3=0.01\,gm/cm^3$
Multiple choice geography crafts and industries development of industries in india traditional crafts and decline of indian industries impact of british rule on kerala

The percentage of carbon in wootz steel is _____.

  1. 3.5

  2. 2.5

  3. 1.6

  4. 0.05
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Wootz Steel is produced by a method known in ancient India. The process involved preparation of porous iron, hammering it while hot to release slag, breaking it up and sealing it with wood chips in a clay container, and heating it until the pieces of iron absorbed carbon from the wood and melted. The steel thus produced had a uniform composition of 1–1.6% carbon and could be heated and forged into bars for later use in fashioning articles, such as the famous medieval Damascus swords.

Multiple choice chemistry endothermic and exothermic reactions endothermic reactions what are enthalpy changes energy change in chemical reactions
For per gram reactant, the maximum quantity of ${ N } _{ 2 }$ gas is produced in which of the following thermal decomposition reaction? (Given: Atomic wt. Cr=52 u, Ba=137 u)
  1. $(NH _4) _2Cr _2O _7(s) \rightarrow N _2(g) + 4H _2O(g)+Cr _2O _3(s)$
  2. $2NH _4NO _3(s) \rightarrow 2N _2(g) + 4H _2O(g)+O _2(g)$
  3. $Ba(N _3) _2(s) \rightarrow Ba(s) + 3N _2(g)$
  4. $2NH _3(g) \rightarrow N _2(g) + 3H _2(g)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

A hollow iron pipe of $21 cm$ long and its external diameter is $8 cm$. If the thickness of the pipes is $1 cm$ and iron weights $\displaystyle 8g/cm^{2}$, then the weight of the pipe is equal to

  1. $3.6 kg$
  2. $3.696 kg$
  3. $36 kg$
  4. $36.9 kg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let external diameter be ${ d } _{ 2 }$ and internal be ${ d } _{ 1 }$
$\Rightarrow { d } _{ 2 }=8cm$ & ${ d } _{ 1 }=8-1-1=6cm$
Therefore, ${ r } _{ 1 }=3cm$ & ${ r } _{ 2 }=4cm$
Volume of hollow cylindrical pipe$=\pi \left( { r } _{ 2 }^{ 2 }-{ r } _{ 1 }^{ 2 } \right) \times h$
$=\cfrac { 22 }{ 7 } \left( { \left( 4 \right)  }^{ 2 }-{ \left( 3 \right)  }^{ 2 } \right) \times 21$
$=\cfrac { 22 }{ 7 } \times 7\times 21$
$=462cm^3$
$\therefore $Weight of the pipe$=8\times 462=3696g=3.696kg$