Chemistry

Mole Concept and Stoichiometry

365 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 evs water: a precious resource importance of water let's play with water water and its properties water resources and conservation

For $\displaystyle 4{ NH } _{ 3 }\left( g \right) +5{ O } _{ 2 }\left( g \right) \rightarrow 4NO\left( g \right) +6{ H } _{ 2 }O\left( g \right) $, if you begin with 16.00 g ammonia and excess oxygen, how many grams of water will be obtained?

  1. 2.294

  2. 36.51

  3. 1.409

  4. 25.3

  5. 2.513
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The molecular weights of ammonia and water are 17 g/mole and 18 g/mole respectively.
4 moles of ammonia gives 6 moles of water.
$4 \times 17 = 68$ g of ammonia will give $6 \times 18 = 108$ g of water.
16.00 g of ammonia will give $16.00 \times \dfrac {108}{68} = 25.3$ g of water.

Multiple choice evs water: a precious resource importance of water let's play with water water and its properties water resources and conservation

For $\displaystyle 4{ NH } _{ 3 }\left( g \right) +5{ O } _{ 2 }\left( g \right) \rightarrow 4NO\left( g \right) +6{ H } _{ 2 }O\left( g \right) $, if you begin with 66.00 g ammonia and 54.00 g oxygen, how many grams of water will be obtained?

  1. 2.294

  2. 36.51

  3. 1.409

  4. 25.3

  5. 2.513
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The molecular weights of ammonia, oxygen and water are 17 g/mole, 32 g/mole and 18 g/mole respectively.

66.00 g ammonia $= \dfrac {66.00}{17} = 3.88$ moles ammonia.
54.00 g oxygen $=\dfrac {54.00}{32.0}=1.6875$ moles oxygen.
4 moles ammonia reacts with 5 moles oxygen.
Hence, 3.88 moles ammonia will react with $3.88 \times \dfrac {5}{4} = 4.85$ moles oxygen but only 1.6875 moles oxygen are present. Hence oxygen is the limiting reagent.

5 moles of oxygen gives 6 moles of water.
1.728 moles of oxygen will give $1.6875 \times \dfrac {6}{5} = 2.025$ moles of water.
The mass of water obtained is $18 \times 2.025 = 36.51$ g of water.

Multiple choice evs water: a precious resource importance of water let's play with water water and its properties water resources and conservation

How many grams of water can be produced when 8 g of hydrogen react with 8 g oxygen?

  1. 8 g

  2. 9 g

  3. 18 g

  4. 27 g

  5. 30 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The molecular masses of hydrogen, oxygen and water are 2 g/mole, 32 g/mole and 18 g/mole respectively.
$2H _2 + O _2 \rightarrow 2H _2O$
 $2 \times 2 = 4$ g of hydrogen reacts with  $1 \times 32 = 32$ g of oxygen to form  $2 \times 18 = 36$ g of water.
Hence, $\dfrac {4}{4} = 1$ g of hydrogen reacts with  $\dfrac {32}{4}  = 8$ g of oxygen to form  $\dfrac {36}{4}= 9$ g of water.

9 grams of water can be produced when 8 g of hydrogen reacts with 8 g oxygen.
Multiple choice chemistry lattice energy born-haber cycle born-haber cycles energy cycles

The Born Haber cycle below represents the energy changes occurring at 298K when $KH$ is formed from its elements
v : ${ \Delta H } _{ atomisation }$ $K = 90 kJ/mol$
w : ${ \Delta H } _{ ionisation }$ $K = 418 kJ/mol$
x : ${ \Delta H } _{ dissociation }$ $H = 436 kJ/mol$
y : ${ \Delta H } _{ electron affinity }$ $H = 78 kJ/mol$
z : ${ \Delta H } _{ lattice }$ $KH = 710 kJ/mol$
On complete reaction with water, $0.1 g$ of $KH$ gave a solution requiring 25 ${ cm }^{ 3 }$ of 0.1M $HCl$ for neutralisation.Calculate the relative atomic mass of potassium from this information.

  1. $39$
  2. $40$
  3. $41$

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

Meq. of KH = Meq. of $HCl$
$\frac { { 0.1 } }{ { E } _{ KH } } \times 1000\quad =\quad 25\quad \times \quad 0.1$
Valency factor (05955) of $K$ is 1 hence
${ E } _{ K }$=${ M } _{ K }$                    ${ M } _{ K }$=39
${ E } _{ KH }$=40                                      ${ E } _{ KH }$=${ E } _{ K }$=
40=${ E } _{ K }$+1                                     ${ E } _{ KH }$
${ E } _{ K }$ $\Rightarrow $ 39

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

The density of aluminium is 2.7 $ \displaystyle g/cm^{3} $. Its density in $ \displaystyle kg/m^{3} $ will be :

  1. 27 $ \displaystyle kg/m^{3} $
  2. 2700 $ \displaystyle kg/m^{3} $
  3. 270 $ \displaystyle kg/m^{3} $
  4. 27000 $ \displaystyle kg/m^{3} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

density of aluminium = $2.7g/{ cm }^{ 3 }$

                                    = $\dfrac { 2.7\times { 10 }^{ -3 }kg }{ { 10 }^{ -6 }{ m }^{ 3 } } $
density of aluminium = $2700kg/{ m }^{ 3 }$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weights $0.25\ kg$ in air and is observe to displace $0.015$ litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, $19.3$ and $8.9$. The approximate percentage of copper in the ornament is

  1. $5\%$
  2. $10\%$
  3. $15\%$
  4. $25\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let volume of gold in ornament $V _1$ and that of copper $=V _2$ and density of gold $=\rho g$ and that of copper$=\rho _c$

$\Rightarrow \rho g V _1+\rho _cV-2=0.25\rightarrow (1)$
Volume of ornament $=$Volume of water displaced
$\Rightarrow V _1+V-2=0.15\times10^{-3}ms\Rightarrow V _2=(0.015\times10^{-3}-V _1)$
According to question-
$\cfrac{\rho _g}{\rho _w}=19.3$ and $\cfrac{\rho _c}{\rho _w}=8.9\ \rho _g=19.3\times10^3 kg/m^3$
$\rho _c=8.9\times10^3 kg/m^3$
Putting these value in equation $(1)$
$(19.3\times 10^{ 3 })V _{ 1 }+(1.9\times 10^{ 3 })(0.0015\times 10^{ -3 }-V _{ 1 })0.25\ [(19.3\times 10^{ 3 })-(1.9\times 10^{ 3 })]V _{ 1 }+(8.9\times 0.015)=0.25\ 10.4\times 10^{ 3 }V _{ 1 }=0.12\ V _{ 1 }=\cfrac { 0.12 }{ 10.4\times 10^{ 3 } } =0.0115\times 10^{ -3 }=1.15\times 10^{ -5 }m^{ 3 }$
$V _2=(0.015\times10^3-0.0115\times10^{-3})=0.0035\times10^{-3}\ \% \quad of\quad copper=(\cfrac{V _2}{V _1+V _2})\times100=[\cfrac{0.0035\times10^{-3}}{(0.0115+0.0035)\times10^{-3}}]\times100$
$\approx 23.34\%\ \approx 25\%$

Multiple choice chemistry substances in common use percent composition water of crystallisation percentage composition and empirical formula

In $FeSO _4$.$7H _2O$, $H _2O$ represents:

  1. molar mass

  2. molecular weight

  3. molar weight

  4. water of crystallisation
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Water of crystallization or water of hydration is the water present inside the crystal. Solids are purified by the procedure called as crystallization in which after purification some pure form of water trap down inside the crystal called as water of crystallization.

Multiple choice chemistry substances in common use percent composition water of crystallisation percentage composition and empirical formula

The number of molecules of water of crystallisation present in one molecule of ferrous sulphate is_______.

  1. $5$
  2. $7$
  3. $6$
  4. $10$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When water is used in the formation of crystals knows as water of crystallization. 'OR' 

Water of crystallization is the total weight of water in a substance at a given temperature and is mostly present in a definite ratio. 
On heating, iron(II) sulfate first loses its water of crystallization and the original green crystals are converted into a brown-colored anhydrous solid. 


$FeSO _4.7H _2O$ the heptahydrate in solution (water as solvent) transforms to both heptahydrate and tetrahydrate when the temperature reaches $56.6 ^\circ C$.


Hence, option $B$ is correct.

Multiple choice chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

What is the percent of composition of the compound that forms when $222.7g$ of N combines compleletly with $77.4g$ of O?

  1. $70%, 30%$
  2. $80%, 20%$
  3. $74.2%, 25.8%$
  4. $72.2%, 27.8%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mass composition$:$
Total mass$: 222.7+77.4=300.1$
Mass percentage of N$:$
$222.7300.1=74.25$
Mass percentage of O$:$
$100-74.2=25.8%$

Multiple choice chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

The formula for % composition of a compound is:

  1. molar mass of a compound/mass due to specific component $\times$ 100
  2. mass due to specific component $\times$ 100
  3. mass due to specific component/molar mass of a compound $\times$ 100
  4. molar mass of a specific component/temperature $\times$ 100
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To calculate the per cent composition of a component in a compound: Find the molar mass of the compound by adding up the masses of each atom in the compound using the periodic table or a molecular mass calculator. Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms. Divide the mass due to the component by the total molar mass of the compound and multiply by 100.

% $\text{composition of a compound}=\cfrac{\text{ Mass due to specific component}}{\text{molar mass of a compound}}\times 100$

Hence, the correct option is $\text{C}$

Multiple choice chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

% composition requires ................ of the compound :

  1. molar mass

  2. temperature

  3. atmospheric pressure

  4. both a and b
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The present composition (by mass) of a compound can be calculated by dividing the mass of each element by the total mass of the compound. i.e. by calculating the $molar$ $mass$ of the compound.

Multiple choice salts salts and their classification acids, bases and salts chemistry

The "alum" used in cooking is potassium aluminum sulfate hydrate, $KAl(SO _{3}) _{2}\cdot xH _{2}O$. To find the value of x, a sample of the compound is heated. The mass of the empty crucible is $20.01\ g$.
The alum hydrate was added to the crucible until the total mass of the crucible and hydrate was $24.75\ g$. The sample was heated in the crucible until the final mass of the crucible and anhydrous product was $22.5\ g$.
What is the value of x?

  1. $2$
  2. $3$
  3. $12$
  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass of crucible Hydrate $=24.75$

Mass of hydrate $=24.75-20.01=4.74$ $g$
Mass of anhydrate $=22.5-20.01=2.49$ $g$
$KAl(SO _3) _2\cdot xH _2O\longrightarrow KAl(SO _3) _2+xH _2O$
Mass: $(18x+226)$ $g$          Mass $=226$ $g$
$(226+18x)$ $g$ $\overset {gives}{\longrightarrow}$ $226$ $g$
$4.74$ $g$ $\overset {gives}{\longrightarrow}$ $2.49$ $g$
$\cfrac{226+18x}{4.74g}=\cfrac{226}{2.49}$
$x\approx 12$