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Test: Stoichiometry - JEE MCQ


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30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Stoichiometry

Test: Stoichiometry for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The Test: Stoichiometry questions and answers have been prepared according to the JEE exam syllabus.The Test: Stoichiometry MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Stoichiometry below.
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Test: Stoichiometry - Question 1
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For the reaction         2x + 3y + 4z → 5w

Initially if 1 mole of x, 3 mole of y and 4 mole of z is taken. If 1.25 mole of w is obtained then % yield of this reaction is

Detailed Solution for Test: Stoichiometry - Question 1

                                 2x + 3y + 4z     =      5w
Moles                        1      3       4
Stoichiometric coff.    2      1       1
Ratio                         0.5    1        1
So, x is Limiting reagent. 
  So 0.5 = moles of w/5  (theoritical)
    Or moles of w = 2.5
But obtained moles= 1.25
Therefore % yield = 1.25/2.5* 100 = 50%

Test: Stoichiometry - Question 2
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A solution of A (MM = 20 ) and B (MM =10), [Mole fraction XB = 0.6] having density 0.7 gm/ml then molarity and molality of B in this solution will be _____ and _____ respectively.

Detailed Solution for Test: Stoichiometry - Question 2

The correct answer is Option A.

Given, molar mass of A, MA = 20g/mol.
and molar mass of B, MB = 10g/mol.
density of solution , d = 0.7 g/ml
mole fraction of B, xB = 0.6
so, mole fraction of A , xA = 1 - 0.6 = 0.4
using formula, 
= 0.6 × 0.7 × 1000/(20 × 0.4 + 10 × 0.6)
= 420/14
= 30 molar
Hence, molarity of the solution is 30M
use formula, 
where m is the molality of the solution.
so, m = 600/(20 × 0.4)
= 600/8
= 75 molal
Hence, molality of the solution is 75m
 

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Test: Stoichiometry - Question 3
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125ml of 8% w/w NaOH solution (sp. gravity 1) is added to 125 ml of 10% w/v HCl solution. The nature of resultant solution would be _____

Detailed Solution for Test: Stoichiometry - Question 3

In 125 ml of 10% NaOH solution.
The mass of NaOH will be,
Mass% = (Mass of NaOH/Volume of solution)*100
= 10% = (Mass of NaOH/125)*100
Mass of NaOH = 12.5g
Mass of NaOH = 12.5/40 
= 0.3125 mole
In 125 ml of 10% HCl solution.
The mass of HCl will be,
Mass% = (Mass of HCl/Volume of solution)*100
= 10% = (Mass of HCl/125)*100
Mass of NaOH = 12.5g
Mass of NaOH = 12.5/36.5 
= 0.3424 mole
The balanced chemical reaction will be,
NaOH + HCl ----> NaCl + H2O
From balanced reaction we conclude that,
As, 1 mole of NaOH will neutralize 1 mole of HCl
So, 0.3125 moles will neutralize 0.3125 moles of HCl
Thus, the moles of HCl left un-neutralized = 0.3424 - 0.3125 = 0.0299 mole
That means, the moles of acid are still present in the solution after mixing. So, the solution will become acidic is nature.

Test: Stoichiometry - Question 4
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Ratio of masses of H2SO4 and Al2(SO4)3 is grams each containing 32 grams of S is _____
Detailed Solution for Test: Stoichiometry - Question 4

Molar mass of the Aluminium sulphate = 342 g/mole.
Mass of the sulphur present in 1 mole of the compound = 32 × 3 = 96 g.
Since, 96 g of S is present in 342 g of compound.
Thus, 32 g of S is present in 114 g of the compound.
Thus, the mass of the Aluminium sulphate is 114 g.
Now,
Molar mass of Sulphuric Acid = 98 g/mole.
Mass of sulphur in 1 mole = 32 g.
Since, 32 g of S is present in 98 g of Acid.
Thus, 32 g of the S is present in 98 g of the Acid.
Now, for the ratio,
Mass of Acid/Mass of Aluminium Sulphate = 98/114
= 49/54 = 0.86

Test: Stoichiometry - Question 5
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The vapour density of a mixture of gas A (Molecular mass = 40) and gas B (Molecular mass = 80) is 25. Then mole % of gas B in the mixture would be
Detailed Solution for Test: Stoichiometry - Question 5

Test: Stoichiometry - Question 6
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For the reaction                    2A + 3B + 5C → 3D

Initially if 2 mole of A, 4 mole of B and 6 mole of C is taken, with 25% yield, moles of D which can be produced are _________
Detailed Solution for Test: Stoichiometry - Question 6

2A + 3B + 5C → 3D
2    4     6    
From given data, it can be seen that A is the limiting reagent.
2 mol of A will produce 3 mol of D. As yield of reaction is 25% only so, mol of D produced
= 3×0.25
= 0.75 mol.
 

Test: Stoichiometry - Question 7
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Two elements X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is :
Detailed Solution for Test: Stoichiometry - Question 7

Let the total mass be 100 gm.
This includes 75.8 gm of X and 24.2 gm of Y
Moles of X = 75.8/75 ∼ 1
Moles of Y = 24.2/16 = 3/2
Therefore, X:Y = 1:3/2 = 2:3
So the formula is X2Y3.

Test: Stoichiometry - Question 8
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Equal volumes of 10% (v/v) of HCl is mixed with 10% (v/v) NaOH solution. If density of pure NaOH is 1.5 times that of pure HCl then the resultant solution be :
Detailed Solution for Test: Stoichiometry - Question 8

Since the density of pure NaOH is 1.5 times that of pure HCl and the volumes (and volume percent) are equal, the mass (and moles) of NaOH will be 1.5 times that of HCl.
NaOH is in excess. 
Therefore, the resultant solution will be basic.

Test: Stoichiometry - Question 9
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10 ml of a compound containing 'N' & 'O' is mixed with 30 ml of H2 to produce H20(l) and 10 ml of N2(g). Molecular formula of compound if both reactants react completely, is
Detailed Solution for Test: Stoichiometry - Question 9

At STP 1 mole of any gas occupies 22.4 L. Hence, the volume of the gases can be correlated to the number of moles of gases. 10 ml of compound reacts with 30 ml of hydrogen to produce water and 10 ml of nitrogen. Thus, for every one mole of the compound, 3 moles of hydrogen reacts to produce water and 1 mole of nitrogen.

X+3H2→H2O+N2
This equation will be balanced when the formula of the compound is , 

 

N2O3, N2O3+3H2→3H2O+N2

Test: Stoichiometry - Question 10
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The percentage by mole of NO2 in a mixture NO2(g) and NO (g) having average molecular mass 34 is :
Detailed Solution for Test: Stoichiometry - Question 10

Test: Stoichiometry - Question 11
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Assuming complete precipitation of AgCl, calculate the sum of the molar concentration of all the ions if 2 lit of 2M Ag2SO4 is mixed with 4 lit of 1 M NaCl solution is :
Detailed Solution for Test: Stoichiometry - Question 11

Moles of Ag2 SO4 = 2M x 2L = 4 mol 
Moles of NaCl = 1M x 4L = 4 mol 

Test: Stoichiometry - Question 12
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What volumes should you mix of 0.2M NaCl and 0.1M CaCl2 solution soo that in resulting solution the concentration of positive ion is 40% lesser than concentration of negative ion. Assuming total volume of solution 1000ml.
Detailed Solution for Test: Stoichiometry - Question 12

Molarity of NaCl, M1 = 0.2 M
Molarity of CaCl2, M2 = 0.1 M
Total Volume of mixture, V = 1000 mL
Let volume of NaCl, V1 = x
Volume of CaCl2 = 1000-x
Since concentration  of cation is 40% less than anion, for every 100 anions contain 60 cation.
NaCl → Na+  +  Cl-
CaCl2 → Ca2+ + 2Cl-
Moles of cation, n1 = 1×M1V  + 1×M2V2
= 0.2 x x + 0.1(1000-x)
= (0.1x + 100) moles
Moles of anion, n2 = 1×M1V1 + 2×M2V2
= 1×0.2 x x + 2×0.1(1000-x)
200 moles
According to question, 
n2-n1/n2 × 100 = 40%
200-(0.1x +100)/200 × 100 = 40
100 - 0.1x = 80
x = 200
Therefore, volume of NaCl, V1 = 200 mL
Volume of CaCl2, V2 = (1000-200) = 800 mL

Test: Stoichiometry - Question 13
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Weight of oxygen in Fe2O3 and FeO is in the simple ratio for the same amount of iron is :
Detailed Solution for Test: Stoichiometry - Question 13

Molar mass of Fe = 56 gm
Molar mass of O = 16 gm
In Fe2O3, mass of Fe = 56 × 2 = 112
Mass of O in Fe2O3 = 48
In FeO, mass of Fe = 56. Since we need to have the same mass of Fe, we need to  take 2 moles of FeO.
Mass of O in 2 moles of FeO = 16 2 =32
Therefore, ratio of mass of O in Fe2O3 and FeO = 48/32 = 3:2

Test: Stoichiometry - Question 14
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The number of atoms present in 0.5 g-atoms of nitrogen is same as the atoms in
Detailed Solution for Test: Stoichiometry - Question 14

Atomic mass of nitrogen = 14 g

Therefore, mass of 1 mole of nitrogen = 14g

Now, 1 mole = 6.022 x 1023 atoms

Since, 1 g-atom = 1 mole

0.5 g-atom = 0.5 moles

Therefore, no. of atoms in 0.5 moles of Nitrogen = 6.022 x 1023 x 0.5

  = 3.011 x 1023

Now,

12 g carbon = 1 mole carbon

32 g sulphur = 1 mole sulphur

8 g oxygen = 0.5 mole oxygen

24 g magnesium = 1 mole magnesium

Number of atoms in 0.5 mole oxygen = = 3.011 x 1023

Test: Stoichiometry - Question 15
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The oxide of a metal contains 30% oxygen by weight. If the atomic ratio of metal and oxygen is 2 : 3, determine the atomic weight of metal.
Detailed Solution for Test: Stoichiometry - Question 15

From the question, formula of oxide = M2O3
%age of oxygen = mass of oxygen in molecule/mass of molecule = 30/100
(16×3)/(2 M+16×3) = 30/100
M = 160-48/2 = 56

Test: Stoichiometry - Question 16
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The O18/O16 ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. The average mass of an atom of oxygen in these meteorites is _____ that of a terrestrial oxygen atom?
Detailed Solution for Test: Stoichiometry - Question 16

Average atomic mass of an element is defined as the summation of the product of the atomic masses and fractional abundances of its isotopes.
Equation to calculate the average atomic mass will be given by:
∑_(i=1 n (Atomic mass of an isotope)i × (fractional abundance)
We are given that the abundance ratio of the oxygen in meteorites is more than the terrestrial oxygen atom.
As, from the formula, average atomic mass of an element is directly proportional to the fractional abundance of the isotopes. Hence, with increase in the abundance, the average atomic mass also increases.
Therefore, the average mass of the oxygen atom in these meteorites will be greater than the terrestrial oxygen atom.

Test: Stoichiometry - Question 17
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Mass of one atom of the element A is 3.9854 x 10-23 . How many atoms are contained in 1g of the element A?
Detailed Solution for Test: Stoichiometry - Question 17

In any element or substance, the number of atoms contained in 1 mole is 6.023 × 10²³ atoms. This is Avogadro's number and is constant in 1 mole of every substance.

mass of 1 atom = mass of a mole of atoms / 6.022 x 10²³

That is molar mass = Avogadro's number × mass of 1 atom

molar mass for A = 6.022 x 10²³ × 3.9854×10⁻²³
                            = 24

Molar mass of element A is 25g/mol


Find number of atoms in 1 gram of element:

If molar mass is 24 g/mol means 1 mole has 24g.

If 24g = 1 mole = 6.022 x 10²³ atoms

Then 1g = 6.022 x 10²³ × 1g/24g

              = 2.5091667 ×10²² atoms

Therefore  1g of element A has 2.5091667 ×10²² atoms

Test: Stoichiometry - Question 18
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Which has the maximum number of atoms of oxygen
Detailed Solution for Test: Stoichiometry - Question 18

Actually 1 mole of H2O contains Avogadro number of H2O molecules. But since each H2O molecule has only 1 Oxygen atom, we say that
number of H2O molecules = number of oxygen atoms
We may also say that the number of Oxygen atoms is directly proportional to the product of number of moles and number of Oxygen atoms in the molecule.
(a) 1 mole of H2O has 22,400 mL
Moles of H2O in 10 mL = 10/22400 = 4.46 10 -4
Moles of O = 4.46×10-4
(b) moles of O = 0.1 x 5 = 0.5
(c) number of moles of O3 = 12/48 = 0.25
No of moles of O =  0.25 x 3 = 0.75
(d) 6 x 1023 molecules →1 mole
12 x 1022 molecules → 0.2 mole
Moles of O = 0.2 x 2 = 0.4
Therefore the maximum number of oxygen atoms are present in 0.1 moles of √V2O5

Test: Stoichiometry - Question 19
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The density of quartz mineral was determined by adding a weighed piece to a graduated cylinder containing 51.2ml water. After the quartz was submersed, the water level was 65.7 ml. The quartz piece weighed  38.4g. What was the density of quartz?
Detailed Solution for Test: Stoichiometry - Question 19

Density=mass/volume. 
Therefore density of quartz =mass /change in volume of water  
Change in volume gives volume of quartz.
Density=38.4/65.7-51.2
=38.4/14.5
=2.648~2.65

Test: Stoichiometry - Question 20
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A sample of clay contains 40% sillica and 15% water. The sample is partially dried by which it loses 5 gm water. If the percentage of water in the partially dried clay is 8, calculate the percentage of silica in the partially dried clay.
Detailed Solution for Test: Stoichiometry - Question 20

 

Clay original sample contains 40% silica , 15% water so total non water component is 100 - 15 = 85 gram .Now sample is partially dried and loses 5 gram water .Again dried clay contains 8 % water means 8 g water . Therefore total non water component is 100-8 = 92 gram.Now percent of silica in partially dried sample will be 40/85×92= 43.29%

Test: Stoichiometry - Question 21
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A definite amount of gaseous hydrocarbon was burnt with just sufficient amount of O2. The volume of all reactants was 600 ml, after the explosion the volume of the products [CO2(g) and H2O(g)] was found to be 700 ml under the similar conditions. The molecular formula of the compound is :
Detailed Solution for Test: Stoichiometry - Question 21

Let the molecular formula of the compound be CmHn.
600 ml of reactants after reaction gives 700 ml of products.

The number of moles are directly proportional to the volume of the gas.

When equations (2) and (3) are solved, m=3 and n=8
Hence, the molecular formula of the compound is C3H8.

Test: Stoichiometry - Question 22
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One gram is silver salt of an organic dibasic acid yields , on strong heating , 0.5934 g of silver. if the weight percentage of carbon in it 8 times the weight percentage of hydrogen and half the weight percentage of oxygen, determine the molecular formula of the acid. [Atomic weight of Ag = 108]
Detailed Solution for Test: Stoichiometry - Question 22

H2A → Ag2A → Ag
Mass of compound = 1 gm
Mass of Ag in compound = 0.5934 gm
Since same mole of Ag be on both sides, 
2 /216+MA = 0.5934/108
Or MA = 148
Therefore molar mass of acid(H2A) = 148+2 = 150gm
Let %age of C = x

Therefore, empirical formula = C2H3O3
(Empirical mass) n = molar mass
n = 150/75 =2
Therefore, molecular formula = C4H6O6

Test: Stoichiometry - Question 23
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C6H5OH(g) + O2(g) → CO2(g) + H2O(/)
Magnitude of volume change if 30 ml of C6H5OH (g) is burnt with excess amount of oxygen, is
Detailed Solution for Test: Stoichiometry - Question 23

The reaction is not balanced. The balanced reaction is 
C6H5OH (g) + 7O2(g)  →  6CO2(g) + 3H2O(l)
Sinec water formed in liquid state, imts volume is considered to be 0.
Volume of C6H5OH = 30 mL
Since 1 volume of C6H5OH reacts with 7 volume of O2,
Volume of O2 = 7×30 = 210mL
7 moles of O2 forms 6 moles of CO2
1 moles of O2 forms 6/7 volume of CO2
210 mL forms 6/7 210 = 180mL
Volume change = (total volume)reactant - (total volume)product
= (30+210)-(180+0) = 60 mL

Test: Stoichiometry - Question 24
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Similar of % labelling of oleum, a mixture of H3PO4 and P4O10 is labelling as ( 100 + x ) % where x  is the maximum mass of water which can react with P4O10 present in 100 gm mixture of H3PO4 and P4O10. If such a mixture is labelled as 127% Mass of P4O10 is 100 gm of mixture, is
Detailed Solution for Test: Stoichiometry - Question 24

P4O10 + 6 H2O --------> 4 H3PO4

(284 gm + 6*18               = 108 gm)

mass of P4O10 = (127 – 100)*284 / 108 = 71 gm

Test: Stoichiometry - Question 25
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Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of hydrogen and 56 lit.O2 at 1 atm & 273 K according to given reaction, is C(s) + H 2(g) + O2(g) → C12H22O11(s)
Detailed Solution for Test: Stoichiometry - Question 25


Test: Stoichiometry - Question 26
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If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is
Detailed Solution for Test: Stoichiometry - Question 26

18 % ⇒ 100 g sample uses 18 g water
⇒ 50 g sample need 9 g water
Since we have added 18 gm water. But according to the question, only 9 gm of water is needed.
⇒ So we have 59 g H2SO4 + 9 g water

Test: Stoichiometry - Question 27
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In the quantitative determination of nitrogen using Duma’s method, N2 gas liberated from 0.42 gm of a sample of organic compound was collected over water. If the volume of N2 gas collected was 100/11 ml at total pressure 860 mm Hg at 250 K, % by mass of nitrogen in the organic compound is
[Aq. tension at 250K is 24 mm Hg and R = 0.08 L atm mol–1 K–1]
Detailed Solution for Test: Stoichiometry - Question 27

The volume of dry nitrogen at STP is 

Percentage by mass of nitrogen 

Test: Stoichiometry - Question 28
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40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution. The liberated CO2 occupied 12.315 lit. at 1 atm & 300 K. The correct option is
Detailed Solution for Test: Stoichiometry - Question 28

To calculate the volume of CO2 at STP condition,
Let say it will be V1
1×V / 273 = 1×12.315 / 300
​⇒V1=11.2L at STP
11.2L of CO2 at STP is equivalent to 0.5 mole of CO2. Suppose, Li is the metal
∴Li2CO3+2HCl⟶2LiCl+H2O+CO2
​∴  0.5 mole of Li2CO3 corresponds to 0.5×73.89 i.e. 37gm of Li2CO3
​∴  Wt. of impurity will be (40−37)gm i.e. 3gm
= 36.5×1 / 0.73=50

Test: Stoichiometry - Question 29
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The minimum mass of mixture of A2 and B4 required to produce at least 1 kg of each product is:
(Given At. mass of ‘A’ = 10; At mass of ‘B’ = 120) 5A2 + 2B4 → 2AB2 + 4A2B
Detailed Solution for Test: Stoichiometry - Question 29

Molar masses of AB2 and A2B are 250 g/mol and 140 g/mole respectively.
1 kg of AB2 corresponds to 4 moles. It requires 10 mole A2 and 4 mole B4 respectively.
1 kg of A2B corresponds to 7.14 mole. It requires 8.9 mole A2 and 3.57 mole B4 respectively.
Thus 10 mole A2 (200 g) and 4 mole B4 (1920  g) will produce at least 1 kg of each product.
The minimum mass of mixture required is 200g+1920g=2120g.

Test: Stoichiometry - Question 30
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The mass of CO2 produced from 620 gm mixture of C2H4O2 & O2, prepared to produce maximum energy is
Detailed Solution for Test: Stoichiometry - Question 30

We know when a fuel undergoes complete combustion it produces carbon dioxide and water. 

The combustion reaction is:

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