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.

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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 botany strategies for enhancement in food production balanced diet food production and microbes plant breeding for food production

250 g of $Methylophilus$ can be expected to produce ______ tonnes of proteins.

  1. 15

  2. 25

  3. 40

  4. 50
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A 250 kg cow produces 200 g of proteins per day. In the same period, 250 g of a microorganisms like Methylophilus methylotrophus because of its high rate of biomass production and growth, can be expected to produce 25 tonnes of protein.

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

$10g$ of limestone on heating produces $4.2g$ of $CaO$. the percentage purity of $Ca{ CO } _{ 3 }$ in limestone is: 

[Atomic mass of $Ca =$ $40$]

  1. $85%$
  2. $75%$
  3. $95%$
  4. $80%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$10g$ of limestone i.e. $CaCO _3$ contains= $\cfrac {10g}{100g/mole}$ moles of $CaCO _3=0.1$ moles of $CaCO _3$

$CaCO _3 \longrightarrow CaO+CO _2$
$1$ mole of $CaCO _3$ produce $1$ mole of $CaO$
Thus $0.1$ moles of $CaCO _3$ must produce $0.1$ mole of $CaO$
$10g$ of $CaCO _3$ must produce $0.1 \times 56= 5.6g$ of $CaO$
But $CaO$ produce is $4.2g$
Pure product obtained is $4.2g$ from $10g$ of $CaCO _3$
Product that obtain along with $1$ m purity from $10g$ of $CaCO _3$ is $5.6g$
So, percentage purity= $\cfrac {\text {mass of pure substance obtained}}{\text {mass of impure substance obtained}}\times 100$
% purity= $\cfrac {4.2}{5.6}\times 100= 75$%

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

A sample contain $Fe\left (SO _{4}  \right ) _{3},$ $FeSO _{4a}$ and impurities. A 600 g sample contains 48g impurities ans equal moles of $Fe _{2}\left (SO _{4}.  \right )in the % of Fe _{2}\left ( SO _{4} \right ) _{3}$ in the mixture is:

  1. 33.33%

  2. 66.7%

  3. 83.33%

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The question has garbled chemical formulas (FeSO4a, SO4 instead of S, O4). Assuming it means Fe2(SO4)3 and FeSO4: Molar mass of Fe2(SO4)3 = 400 g/mol, FeSO4 = 152 g/mol. For equal moles (let n = 1), mass of Fe2(SO4)3 = 400 g, mass of FeSO4 = 152 g. Total pure = 552 g. % of Fe2(SO4)3 in pure = (400/552) × 100 = 72.5%. In total mixture (600 g): 400/600 = 66.7%. The answer depends on whether we ask % in pure mixture or total sample.

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

$4$ g of hydrogen $(H _2)$, $64$g of sulphur (S) and $44.8$ L of $O _2$ at STP react and form $H _2SO _4$. If $49$g of $H _2SO _4$ is formed, then $\%$ yield is  ?

  1. $25\%$
  2. $50\%$
  3. $75\%$
  4. $100\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$H _2+S+2O _2\longrightarrow H _2SO _4$

 $2g$    $32g$    $\underset {|||}{64g}$           $98g$
                  $44.8L$
The mole ratio is $H:S:O=1:1:2$
Since $O _2$ is limiting only $98g$ of $H _2SO _4$ is formed.
Given $49g$ of $H _2SO _4$ is formed.
So % yield= $\cfrac {49}{98}\times 100=50$% .

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

A metal oxide (MO) is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 7.95 g of oxide requires 0.2 g of $H _2$ to yield 6.35 g of the metal. We may deduce that:

  1. The atomic weight of the metal is 48

  2. The atomic weight of the metal is 16

  3. The atomic weight of the metal is 12

  4. The atomic weight of the metal is 63.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$2MO + 2H _2 \rightarrow 2M + 2H _2O$

7.92 g    0.2g       6.35 g
let metal weight is x
mol  mol conclution 
$\dfrac{0.2}{2}$ = $\dfrac{6.35}{x}$
we get      x  =  63.5  
ans is D

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

20 g of a magnesium carbonate sample decomposes on heating to given carbondioxide, and 8g magnesium oxide. What will be the percentage of purity of ${\text{MgC}}{{\text{O}} _3}$ sample ? 

  1. 96

  2. 60

  3. 84

  4. 75
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

MgCO3 -> MgO + CO2. 8 g MgO = 8/40 = 0.2 mol. This requires 0.2 mol MgCO3 = 0.2 * 84 = 16.8 g. Purity = (16.8 / 20) * 100 = 84%.

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

Consider the reaction:
$2ZnS + 3O _{2}\rightarrow 2ZnO + 2SO _{2}$
This reaction has an $80.0$ yield.
What mass of $ZnO$ is produced when $50.0\ g\ ZnS$ is heated in an open vessel untill no further weight loss is observed?

  1. $33.4\ g$
  2. $40.4\ g$
  3. $43.4\ g$
  4. $3240\ g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Molar mass of $ZnS = 97.5\space g$

No. of moles of $ZnO = \dfrac{50}{97.5} = 0.5128\space moles$
$2\space moles$ of $ZnS$ produce $2\space moles$ of $ZnO.$
So, $0.5128\space moles$ produce $0.5128\space moles$ of $ZnO.$
So, mass of $ZnO = (0.5128)\times 81 = 41\space g$
As percentage yield $=80\%$
$\Rightarrow$ Mass of $ZnO = \dfrac{80}{100} \times 41 = 33.4\space g$

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

$2Ag{ NO } _{ 3 }+Cu\rightarrow Cu{ \left( { NO } _{ 3 } \right)  } _{ 2 }+2Ag$
What is the percent yield when $0.17\ g$ of $Ag{NO} _{3}$ in aqueous solution reacts with excess copper to produce $0.08\ g$ $Ag$? (At. mass of $Ag=107\ g/mol$) 

  1. $74$%
  2. $47$%
  3. $89$%
  4. $65$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ 2AgNO _3 \space + \space Cu \rightarrow Cu(NO _3) _2 \space + \space 2Ag $

Percentage of Ag in $AgNO _3 = \dfrac{108 \times 100}{108 + 14 + 48} = \dfrac{108}{170} \times 100 = \dfrac{1080}{17} = 63.52\%$

So, amount of Ag produced $= \dfrac{63.52}{100} \times 0.17 = 0.108\space g$

$\%$ Yield $= \dfrac{0.08}{0.108} \times 100 = 74\%$