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

Iron pyrites has formula $FeS _2. (Fe = 56; S = 32)$. What is the mass of sulfur contained in 30 grams of pyrites?

  1. 16 g

  2. 32 g

  3. 20 g

  4. 24 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molecular weight of $ \displaystyle FeS _2 =  56+2(32)=120$ g/mol.

One mole (120 g) of $ \displaystyle FeS _2$ will contain $ \displaystyle 2 \times 32 = 64$ g of $S$
30 g of $ \displaystyle FeS _2$ will contain $ \displaystyle \dfrac {30}{120} \times  64=16$ g of $S$.
Hence, the mass of sulfur contained in 30 grams of pyrites is 16 g.

Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept
An organic compound contains 69% carbon, 4.8% hydrogen and the remaining is oxygen. calculate the masses of carbon dioxide produced when 0.20g of substance is subjected to combustion.

  1. 0.40 g

  2. 0.50 g

  3. 0.60 g

  4. 0.70 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Percentage of carbon in organic compound $= 69 \%$

So, 100 g of organic compound contains 69 g of carbon.

∴ 0.2 g of organic compound will contain $=\dfrac{ 69 \times 0.2}{ 100} = 0.138$ g of carbon

The molecular mass of carbon dioxide, $CO _2 = 44$ g

So, 12 g of carbon is contained in 44 g of $CO _2$.

0.138 g of carbon will be contained $= \dfrac{44 \times 0.138}{12} = 0.506$ g of carbon.

Thus, 0.506 g of $CO _2$ will be produced on complete combustion of 0.2 g of an organic compound.
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

The mass of one molecule of water is approximately:

  1. $1\ g$
  2. $0.5\ g$
  3. $1.66 \times 10^{-24}\, g $
  4. $ 3 \times 10^{-23}\, g $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Mass of $6.0 \times 10^{23}$ molecules of water is 18 g.

Mass of 1 molecule of water $=\dfrac{M}{N _A}=\dfrac{18}{6.0\times 10^{23}}$

Mass of 1 molecule of water $=3. \times$ $10^{-23}$ g 
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

If of conservation of mass was to hold true, then 20.8 g of ${ BaCl } _{ 2 }$ on reaction with 9.8 g of ${ H } _{ 2 }{ SO } _{ 4 }$ will produce 7.3 g of $HCl$ and ${ BaSO } _{ 4 }$ equal to :

  1. 11.65 g

  2. 23.3 g

  3. 25.5 g

  4. 30.6 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$BaCl _2+H _2SO _4 \longrightarrow BaSO _4+2HCl$


$1$ mole of $BaCl _2$ reacts with $1$ mole of $H _2SO _4$ to give $1$ mole of $BaSO _4$ and $2$ moles of $HCl$.


Here, moles of $BaCl _2=\dfrac{20.8}{208}=$ moles of $H _2SO _4= \dfrac{9.8}{98}=0.1$

$\therefore$ Moles of $BaSO _4$ formed $=0.1$

$\therefore$ Mass of $BaSO _4$ formed $=0.1 \times 233= 23.3 g$


i.e. $20.8+9.8=7.3+23.3=30.6$

Hence the correct option is B.

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

The number of electrons which will together weigh one gram is :

  1. $1.098 \times 10^{27}$ electrons
  2. $9.1096\times 10^{31}$ electrons

  3. 1 electrons
  4. $1\times 10^4$ electrons
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass of a electrons = $9.1096\times 10^{-31}Kg$
1g or $10^{-3}kg = \dfrac{1}{9.1096\times 10^{-31}}\times 10^{-3}$
=$1.098\times 10^{27}$ electons

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

What is the mass of oxalic acid, $ H _{2}C _{2}O _{4},$ which can be oxidized to $ CO _{2}$ by 100 ml of $MnO _{4}^{-}$ solution, 10 ml of which is capable of oxidizing 50 ml of $ 1.00 N\ I^{-} $ to $ I _{2}?$

  1. 2.25 g

  2. 52.2 g

  3. 25.2 g

  4. 22.5 g
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Balanced chemical reaction,
$2KMnO _{4}+5H _{2}C _{2}O _{4}+3H _{2}SO _{4}\rightarrow 2MnSO _{4}+10CO _{2}+K _{2}SO _{4}+8H _{2}O$

$(KMnO _{4})N _{1}V _{1}= N _{2}V _{2}(I _{2})$

$N _{1}\times 10= 1\times 50$

$N _{1}= 5N$

n-factor for $KMnO _{4}= 7-2=5$

Moles of $KMnO _{4}=\dfrac{5}{5}=1$

2 mole $KMnO _{4}= 5$ mole $H _{2}C _{2}O _{4}$

1 mole $KMnO _{4}= 2.5$ mole $H _{2}C _{2}O _{4}$

In 100 mL or 0.1 L $= 0.1\times 2.5= 0.25$ moles
Mass of $H _{2}C _{2}O _{4}= 0.25\times 90= 22.5g$

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

Two acids $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed, respectively. Find the ratio of the masses of $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ 

  1. $ 1:1 $
  2. $ 1:2 $
  3. $ 2:1 $
  4. $ 2:3 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$H _{2}SO _{4}+2NaOH\rightarrow Na _{2}SO _{4}+2H _{2}O$

$H _{3}PO _{4}+NaOH\rightarrow NaH _{2}PO _{4}+H _{2}O$

Equivalent of alkali $= 19$ eq of $H _{2}SO _{4}= 1g$ eq of $H _{3}PO _{4}$

Two acids must be reacting in the ratio of their equivalent masses

Eq. wt. of $H _{2}SO _{4}=\dfrac{98}{2}=49$

Eq. wt. of $H _{3}PO _{4}= \dfrac{98}{1}=98$

$\therefore $ ratio of masses of $H _{2}SO _{4}$ & $H _{3}PO _{4}$
$49:98=1:2$

$\Rightarrow 1:2$

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

$1.25$ g of sample of limestone on heating gives $0.44$ g carbon dioxide. The percentage purity of $CaCO _3$ in limestone is:

  1. $75\%$
  2. $85\%$
  3. $90\%$
  4. $80\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$CaCO _3$ $\quad \underrightarrow \Delta\quad  CaO +CO _2$

Moles of $CO _2$ produced =$\cfrac{0.44}{44}$$=0.01$ moles
Moles of pure $CaCO$$ _3$ required $=0.01$ moles
$=0.01 \times 100g$
$=1g$
Therefore $\%$ purity of $CaCO$$ _3$  in limestone =$\cfrac{1}{1.25}\times 100$
$=80\%$

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

$50\ g$ of an impure calcium carbonate sample decomposes on heating to give carbon dioxide and $22.4\ g$ calcium oxide. The percentage purity of calcium carbonate in the sample is:

  1. $60\%$
  2. $80\%$
  3. $90\%$
  4. $70\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$CaCO _3\overset { \Delta  }{ \rightleftharpoons  } CaO+CO _2$

Let pure sample of $CaCO _3=x$ $grams$
Mass of $CaO$ produced after decomposition $=22.4$ $g$
Molar mass of $CaCO _3=100$ ${g/mol}$
and, Molar mass of $CaO=56$ $g$
If $100\%$ is pure, then
$100$ $g$ $CaCO _3\longrightarrow 56$ $g$ of $CaO$
Also,$y$ $g$ of $CaCO _3\longrightarrow 22.4$ $g$ of $CaO$
Dividing these two,
$\cfrac{100}{y}=\cfrac{56}{22.4}$
$y=40$ $grams$
$\therefore$ Percentage of purity $=\cfrac{Mass\quad of \quad pure\quad sample}{Total\quad mass\quad of\quad impure}\times 100=\cfrac{40}{50}\times 100=80\%$

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

What is the purity of concentrated $H _2SO _4$ solution $(d=1.8gm/mol)$ if $5\text{ ml}$ of these solution is neutralized by $84.5 \text{ ml}$ of $2N \text{ NaOH}$ solution. 

  1. $93 \text {%}$
  2. $94.6 \text {%}$
  3. $92.12 \text {%}$
  4. $91.5 \text {%}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$H _2SO _4+2NaOH \rightarrow Na _2SO _4+2H _2O$
Moles of $NaOH$ required $=\cfrac {2 \times 84.5}{1000}=0.169$
For $2 \ moles \ NaOH \rightarrow 1 \ mole \ H _2SO _4$ is used
For $\ 0.169 \ mole \ NaOH \rightarrow 0.0845 \ moles$ are used.
Mass of $H _2SO _4 \rightarrow 0.0845 \times 98=8.281 \ gm$ (Theoretical)
Mass of $H _2SO _4$ used in original reaction $\Rightarrow 5 \times 1.8 = 9 \ gm$
$\therefore$ % purity $=\cfrac {8.281}9 \times 100 = 92.12$ %

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

$12.5$ g of an impure sample of limestone on heating gives $4.4$ g of carbon dioxide. The percentage purity of $CaCO _{3}$ in the sample is:

  1. $72$%
  2. $75$%
  3. $80$%
  4. $85$%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ CaCO } _{ 3 }\overset { \Delta  }{ = } CaO+{ CO } _{ 2 }\uparrow $

$100gm$                      $44gm$
Therefore $1$ mole i.e. $100gm$ of ${ CaCO } _{ 3 }$ (lime stone) give $1$ mole i.e. $44gm$ of ${ CO } _{ 2 }$.
$4.4gm$ of ${ CO } _{ 2 }$ are produced from $\dfrac { 100\times 4.4 }{ 44 } gm$ i.e. $10gm$ of ${ CaCO } _{ 3 }$.
$\therefore$   The percentage of purity $=\left( 1-\dfrac { 12.5-10 }{ 12.5 }  \right) \times 100$% $=80$%
$\therefore$   Correct answer is $C$ $(80$%$)$.

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

0.2828 g of iron wire was dissolved in excess of dilute $H _2SO _4$ and the solution was made upto 100 ml. 20 ml of this solution required 30 ml. of $\dfrac{N}{30} K _2Cr _2O _7$ solution for oxidation. Calculate % purity of iron in the wire:

  1. 99

  2. 95

  3. 90

  4. 85
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Meq of K2Cr2O7 = normality (1/30) * volume (30 ml) = 1 meq. This oxidizes 1 meq of Fe2+ in the 20 ml aliquot. In the total 100 ml solution, there are 1 * (100 / 20) = 5 meq of Fe2+. Mass of iron = meq * eq weight / 1000 = 5 * 56 / 1000 = 0.28 g. Percentage purity = (0.28 / 0.2828) * 100 is approximately 99 percent.