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 evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

$2.16$ grams of Cu, on reaction with $HNO _{3}$, followed by ignition of the nitrate, gave $2.7$ g of copper oxide. In another experiment $1.15$ g of copper oxide, upon reaction with hydrogen, gave $0.92$ g of copper. This data illustrate the law of:

  1. multiple proportions

  2. definite proportions

  3. reciprocal proportions

  4. conservation of mass
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the first sample of copper oxide (obtained by the action of nitric acid on Cu), the mass of Cu is 2.16 g and the mass of oxygen is $\displaystyle 2.7 - 2.16  =  0.54$ g. 

The ratio of the mass of Cu to the mass of oxygen is $\displaystyle \dfrac {2.16}{0.54} = 4 :1$

In the second sample of copper oxide (which reacts with hydrogen) , the mass of Cu is 0.92 g and the mass of oxygen is $\displaystyle 1.15 - 0.92  =  0.23$ g. 
The ratio of the mass of Cu to the mass of oxygen is $\displaystyle \dfrac {0.92}{0.23} = 4 :1$
Hence, this illustrates the law of definite Proportions.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

A sample of calcium carbonate $\displaystyle \left ( CaCO _{3} \right )$ has the percentage composition as given: $Ca = 40\%,\ C = 12\%,\ O = 48\%$. 


If the law of constant proportions is true, then the weight of calcium in $4$ g of a sample of calcium carbonate obtained from another source will be :

  1. $0.016$ g
  2. $0.16$ g
  3. $1.6$ g
  4. $16$ g
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In $100 $ g $\displaystyle CaCO _{3}$, weight of $Ca$ is $40 $ g.

In $4$ g $\displaystyle CaCO _{3}$, weight of $Ca$ is $=\displaystyle \frac{40}{100}\times 4=1.6$ g

Hence, the correct option is $C$

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

1.2375 g of cupric oxide on being heated in a current of hydrogen gave 0.9322 g of the metal In another experiment 0.9369 g of pure copper was dissolved in nitric acid Excess of acid evaporated and the residue has ignited The weight of the cupric oxide left was 1.2469 g. Which law of chemical combination is shown by the above results?

  1. Law of constant proportions

  2. Law of multiple proportions

  3. Law of conservation of mass

  4. Law of constant volumes
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

law of multiple proportion is shown as copper reacts with two different compounds .

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

$1.375$ g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was $1.098$ g. In another experiment, $1.156$ g of copper was dissolved in nitric acid and the resulting solution was evaporated to dryness. The residue of copper nitrate when strongly heated was converted into $1.4476$ g of cupric oxide. State the law illustrated by these chemical combinations.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

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

Law of constant composition states that all samples of a given chemical compound have the same elemental composition by mass. 

Here, in the first experiment 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper obtained was 1.098 g.
Which means 1.375 g of cupric oxide has 1.098 g of carbon and $1.375-1.098 = 0.277$ g of oxygen. 
Copper and oxygen are in the ration $\frac{1.098}{0.277} = \frac{1}{4}$. 
In the second experiment 1.156 g of copper was dissolved in nitric acid to form copper nitrate, which on strongly heating got converted into 1.4476 g of cupric oxide. 
Which means 1.4476 g of cupric oxide has 1.156 g of copper and $ 1.4476-1.156 =0.2911 g $ oxygen.
Copper and oxygen are in the ratio $\frac{1.156}{0.2911} = \frac{1}{4}$.
Which mean what ever may be the source of a compound, it has same elemental composition by mass in all samples.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Hydrogen combines with nitrogen in a 3:14 weight ratio to form ammonia. If every molecule of ammonia contains three atoms of hydrogen and one atom of nitrogen, an atom of nitrogen must weigh :

  1. 14 times the mass of a hydrogen atom

  2. 14/3 times the mass of a hydrogen atom

  3. 3 times the mass of a hydrogen atom

  4. 3/14 times the mass of a hydrogen atom
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Atoms of different elements combine in simple whole number ratios to form chemical compounds. Combination always takes place in the simplest possible way between particles of different weights.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Hydrogen combines with oxygen in a 1:8 weight ratio to form water. If every molecule of water contains two atoms of hydrogen and one atom of oxygen, an atom of oxygen must weigh :

  1. 8 times the mass of a hydrogen atom

  2. 16 times the mass of a hydrogen atom

  3. 1/16 times the mass of a hydrogen atom

  4. 1/8 times the mass of a hydrogen atom
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The law of definite proportions and constant composition states that all samples of a given chemical compound have the same elemental composition by mass e.g. oxygen makes up up about 8/9th of mass of any sample of pure water while hydrogen makes up the remaining 1/9th of mass.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

When $3\ g$ of carbon is burnt in $8\ g$ of oxygen, $11 g$ of carbon dioxide is produced. What mass of carbon dioxide will be formed when $3\ g$ of carbon is burnt in $50\ g$ oxygen?

  1. $12\ g$
  2. $13\ g$
  3. $11\ g$
  4. $10\ g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This is governed by the law of definite proportion. Carbon and oxygen combine in $3:8$ ratio. So $11 \ g$ of carbon dioxide is formed. 


Hence, in the latter case too, only $8 \ g$ of $O _2$ will be reacted with 3g of carbon (because carbon is the limiting reagent here ) to form 11g of $CO _2$ and  $42 \ g$ of oxygen will remain unreacted.

Hence, the correct option is C.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

When $1.375\ g$ of cupric oxide is reduced on heating in a current of hydrogen, the weight of copper remaining $1.098\ g$. In another experiment, $1.179\ g$ of copper is dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed is $1.476 \ g$. This is in agreement with :

  1. Law of definite proportion

  2. Law of mass conservation

  3. Law of momentum conservation

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

First experiment
$\bullet$ Copper oxide $=1.375 g$
$\bullet$ Copper left $= 1.098 g$
$\bullet$ Oxygen present $= 1.375 - 1.098 = 0.277 g$
Percentage of oxygen in $CuO = (0.277) (100$ %) $1.375 = 20.15$ %

Second Experiment
$\bullet$ Copper taken $= 1.179 g$
$\bullet$ Copper oxide formed $= 1.476 g$
$\bullet$ Oxygen present $= 1.476 - 1.179 = 0.297 g$
Percentage of oxygen in $CuO = (0.297) (100$ %) $1.476 = 20.12$ %
Percentage of oxygen is approximately (within significant figures limit) the same in both the above cases. So the law of constant composition is illustrated.

In chemistry, the law of definite proportion, sometimes called Proust's law or the law of definite composition, or law of constant composition states that a given chemical compound always contains its component elements in fixed ratio and does not depend on its source and method of preparation.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

A sample of $CaCO _3$ has Ca - 40%, C = 12% and 0 = 48%. If the law of constant proportions is true, then the mass of Ca in 5 g of $CaCO _3$ from another source will be:

  1. 2.0 g

  2. 0.2 g

  3. 0.02 g

  4. 20.0 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since, mass percentage of Ca in $CaCO _3 = 40 \text{%} $ and law of constant proportion is true.
So, the mass percentage of $Ca $ in $ 5gm$ $CaCO _3$ will also be $40\text%.$
Hence,$ 40\text{% of 5gm =}$$ 5\times $$\dfrac{40}{100}$ $= 2gm $
Hence, answer is option A.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Zinc sulphate contains 22.65% Zn and 43.9% $H _2O$. If the law of constant proportions is true, then the mass of zinc required to give 40 g crystals will be:

  1. 90.6 g

  2. 9.06 g

  3. 0.906 g

  4. 906 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ZnSO _4-22.65\% Zn $ $43.9\% H _2O$


$100g$ crystal has $22.65$ $Zn$


$40g$ crystal has $x$ $Zn$

$x=\cfrac{40\times 22.65}{100}=\cfrac{906}{100}=9.06g$
Hence, mass of $Zinc$ required is $9.06g$.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

When one mole each of CO and $O _2$ are made to react at STP, the total number of moles at an end of the reaction is:

  1. 1.5 moles

  2. 1 mole

  3. 4 moles

  4. 2 moles
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2CO + O _2 \rightarrow 2CO _2$


Initially, we have one mole of each CO and $O _2$.

1 mole of CO will react with 0.5 mole of $O _2$ to form 1 mole of $CO _2$.

the total no. of moles at the end of the reaction is 0.5 mole $O _2$ + 1 mole of CO$ _2$= 1.5 moles

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Zinc sulphate contains $22.65$% $Zn$ and $43.9$% ${H} _{2}O$. If the law of constant proportions is true, then the mass of zinc required to give $40g$ crystal will be:

  1. $9.06g$
  2. $90.6g$
  3. $0.906g$
  4. $906g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
100g of crystals are obtained from = 22.65g of $Zn$

$\therefore $ 40 g of crystals are obtained from =$\frac { 22.65 }{ 100 } \times40$

= 9.06g 
Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

$14$ g of element X combines with $16$ g of oxygen. On the basis of this information, which of the following is a correct statement?

  1. The element X could have an atomic weight of $7$ amu and its oxide is $XO$
  2. The element X could have an atomic weight of $14$ amu and its oxide is $\displaystyle X _{2}O$
  3. The element X could have an atomic weight of $7$ amu and its oxide is $\displaystyle X _{2}O$
  4. The element X could have an atomic weight of $14$ amu and its oxide is $\displaystyle XO _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$14$ g of an element X combines with $16$ g of oxygen, then the element X could have an atomic weight of $7$ amu and its oxide is $X _2O $.
When the atomic weight is $7$, $14$ g will corresponds to $2$ moles.
$16$ g of oxygen corresponds to $1$ mole. 

The formula $X _2O $ suggests the valency of $1$ for X.