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 zoology excretion in living organisms excretion - animals human excretory system and products excretory system and products excretion - plants excretion in plants

How many molecules of ammonia are required to form 8 molecules of urea?

  1. 24

  2. 8

  3. 16

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

Urea synthesis is a five-step cyclic process, with five distinctive enzymes. One turn of the cycle consumes 2 molecules of ammonia and  1 molecule of carbon dioxide and creates 1 molecule of urea ((NH2)2CO. Hence  16  molecules of ammonia are required to form 8 molecules of urea.

So, the correct answer is 'Option C'.

Multiple choice c3, c4 and cam plants photosynthesis in higher plants biology

If 24G-3-P molecules are formed in ${ C } _{ 3 }$ plants.Calculate a) the number of $CO _{ 2 }$ used and b) the number of G3P used for regeneration RuBP molecules

  1. $a=12, b=20$
  2. $a24, b=48$
  3. $a=12, b=24$
  4. $a=20, b=12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In carbon fixation 5 molecules of C)@ are combines or fixed with 6 molecules; Ribulose Bisphosphate(RuBP) to make 12 molecules of PGA.

The remaining 20 G3P molecules regenerate RuBP, which enables the system to prepare for the carbon fixation step.
So, the correct option is 'a = 12, b = 20'.

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 botany how plants grow experiments on photosynthesis - light importance of light for photosynthesis photosynthesis necessities- experiments various experiments on photosynthesis experiments on photosynthesis - gases

How many molecules of water are needed by a green plant to produce one molecule of hexose reduce $6$ molecules of $CO _{2}$?

  1. $6$
  2. $12$
  3. $24$

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

The balanced equation for photosynthesis is 6CO2 + 12H2O -> C6H12O6 + 6O2 + 6H2O. To produce one molecule of hexose (C6H12O6) from 6 molecules of CO2, 12 molecules of water are consumed, but 6 are released, resulting in a net requirement of 6 molecules of water.

Multiple choice zoology introduction to microbiology microbial pollution control environmental use of microorganism microbes in waste treatment and biogas

In secondary treatment, solid reducers, ranges from

  1. 70-92%

  2. 80-90%

  3. 20-60%

  4. 10-70%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Secondary treatment typically removes 70-92% of solids and BOD (Biochemical Oxygen Demand) from wastewater. The lower range (70-80%) represents less efficient systems, while well-operated plants achieve 85-92% removal.

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.