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NEET Exam  >  NEET Test  >  NCERT Based Tests  >  Test: Stoichiometry and Stoichiometric Calculations - NEET MCQ

Stoichiometry and Stoichiometric Calculations - Free MCQ Practice Test


MCQ Practice Test & Solutions: Test: Stoichiometry and Stoichiometric Calculations (20 Questions)

You can prepare effectively for NEET NCERT Based Tests for NEET with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Stoichiometry and Stoichiometric Calculations". These 20 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 20

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Test: Stoichiometry and Stoichiometric Calculations - Question 1
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1 g of Mg is burnt in a closed vessel containing 0.5 g of O2. Which reactant is limiting reagent and how much of the excess reactant will be left?

Detailed Solution: Question 1


48 g of Mg requires 32 g of O2
1 g of Mg requires 32/48 = 0.66 g of O2
Oxygen available = 0.5 g
Hence, O2 is limiting reagent.
32 g of O2 reacts with 48 g of Mg
0.5 g of O2 will react with 48/32 x 0.5 = 0.75 g of Mg
Excess of Mg = (1.0 - 0.75) = 0.25 g

Test: Stoichiometry and Stoichiometric Calculations - Question 2
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In a reaction container, 100 g of h2 and 100 g of CI2 are mixed for the formation of HCl gas. What is the limiting reagent and how much HCl is formed in the reaction?

Detailed Solution: Question 2


2 g of H2 reacts with 71 g of CI2
100 g of H2 will react with 71/2 x 100 = 3550g of CI2
Hence, CI2 is the limiting reagent.
71 g of CI2 produces 73 g of HCl
100g of CI2 will produce 73/71 x 100 = 102.8 g of HCl

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Test: Stoichiometry and Stoichiometric Calculations - Question 3
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If 40 g of CaCO3 is treated with 40 g of HCl, which of the reactants will act as limiting reagent?

Detailed Solution: Question 3


100 g of CaCO3 reacts with 73 g of HCl
40 g of CaCO3 will react with 73/100 x 40 = 29.2 g of HCl
Since CaCO3 is completely consumed and some amount (40 - 29.2 = 10.8g) of HCl remains unreacted and hence, CaCO3 is limiting reagent.

Test: Stoichiometry and Stoichiometric Calculations - Question 4
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The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 is

Detailed Solution: Question 4

AgNO3 + NaCl → AgCl + NaNO3
No. of moles of AgNO3 = 3.4/170 = 0.02
No, of moles of NaCl = 5.85/58.5 = 0.1
Limiting reagent = AgNO3
1 mole of AgNO3 produces 1 mole of AgCl
0.02 mole of AgNO3 will produce 0.02 mole of AgCl
Weight of AgCl produced = 0.02 x 143.5 = 2.870 g.

Test: Stoichiometry and Stoichiometric Calculations - Question 5
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How much oxygen is required for complete combustion of 560 g of ethene?

Detailed Solution: Question 5


28 g of C2H4 requires 96 g of O2
560 g of C2H4 requires 96/28 x 560
= 1920 g or 1.92 kg of O2

Test: Stoichiometry and Stoichiometric Calculations - Question 6
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What is the mass percent of oxygen in ethanol?

Detailed Solution: Question 6

Molecular formula of ethanol = C2H5OH
Molar mass of ethanol of 2 x 12.01 + 6 x 1.008 + 16 = 46.068g
Mass percent of oxygen = 16/46.068 x 100 = 34.73%

Test: Stoichiometry and Stoichiometric Calculations - Question 7
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How much mass of sodium acetate is required to make 250 mL of 0.575 molar aqueous solution?

Detailed Solution: Question 7

Molar mass of sodium acetate (CH3COONa)
= 82.0245 g/mol
Mass of CH3COONa required to make 250 mL of 0.575 M solution 

Test: Stoichiometry and Stoichiometric Calculations - Question 8
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How much copper is present in 50 g of CuSO4?

Detailed Solution: Question 8

Molar mass of CuSO4 = 63.5 + 32 + 4 x 16 = 159.5 g
Mass of copper present in 159.5 g of CuSO4 = 63.5 g
∴ Mass of copper present in 50 g of CuSO4

Test: Stoichiometry and Stoichiometric Calculations - Question 9
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A solution is prepared by adding 5 g of a solute 'X' to 45 g of solvent 'Y'. What is the mass per cent of the solute 'X'?

Detailed Solution: Question 9

Mass percent of X

Test: Stoichiometry and Stoichiometric Calculations - Question 10
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An impure sample of silver (1.5 g) is heated with S to form 0.124 g of Ag2S. What was the per cent yield of Ag2S?

Detailed Solution: Question 10

216 g of Ag forms 248 g of Ag2S
1.5 g of Ag forms 248/216 x 1.5 = 1.722 g of Ag2S
% yield of Ag2S = 0.124/1.722 x 100 = 7.2%

Test: Stoichiometry and Stoichiometric Calculations - Question 11
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2.82 g of glucose is dissolved in 30 g of water. The mole fraction of glucose in the solution is

Detailed Solution: Question 11

No. of moles of glucose = 2.82/180 = 0.01567
No. of moles of water = 30/18 = 1.667
Total no. of moles of solution = 0.01567 + 1.667 = 1.683
Mole fraction of glucose = 0.01567/1.683 = 0.0093 = 0.01

Test: Stoichiometry and Stoichiometric Calculations - Question 12
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What volume of water is to be added to 100 cm3 of 0.5 M NaOH solution to make it 0.1 M solution?

Detailed Solution: Question 12

M1V1 = M2V2
0.5 x 100 = 0.1 x V2
V2 = 500 cm3
Volume of water to be added to 100 cm3 of solution
= 500 - 100 = 400 cm3

Test: Stoichiometry and Stoichiometric Calculations - Question 13
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The final molarity of a solution made by mixing 50 mL of 0.5 M HCl, 150 mL of 0.25 M HCl and water to make the volume 250 mL is

Detailed Solution: Question 13

M1V1 = M2V2 = MV
0.5 x 50 + 0.25 x 150 = M x 250
M = (25 + 37.5)/250 = 0.25 M

Test: Stoichiometry and Stoichiometric Calculations - Question 14
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A solution is made by dissolving 49 g of H2SO4 in 250 mL of water. The molarity of the solution prepared is

Detailed Solution: Question 14

Molarity

= 2 M

Test: Stoichiometry and Stoichiometric Calculations - Question 15
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What is the concentration of copper sulphate (in mol L-1) if 80 g of it is dissolved in enough water to make a final volume of 3 L?

Detailed Solution: Question 15

Molar mass of CUSO4 = 63.5 + 32 + 64 = 159.5
Moles of CUSO4 = 80/159.5 = O.50
Volume of solution = 3 L

= 0.167 mol L-1

Test: Stoichiometry and Stoichiometric Calculations - Question 16
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4.28 g of NaOH is dissolved inwater and the solution is made to 250 cc. What will be the molarity of the solution?

Detailed Solution: Question 16

No. of moles of NaOH = 4.28/40 = 0.107 Volume of solution = 250 cm3
M = n/V(in L) = 0.107/250 x 1000 = 0.428 mol L-1

Test: Stoichiometry and Stoichiometric Calculations - Question 17
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What volume of 5 M Na2SO4 must be added to 25 mL of 1 M BaCl2 to produce 10 g of BaSO4?

Detailed Solution: Question 17

Na2SO4 + BaCI2 → BaSO4 + 2NaCl
No. of moles of BaSO4 = w/M = 10/233 = 0.0429
∴ No. of moles of Na2SO4 needed = M x V/1000
Or 0.0429 = 5 x V/1000
V = 8.58 mL

Test: Stoichiometry and Stoichiometric Calculations - Question 18
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What will be the molarity of the solution in which 0.365 g of HCl gas is dissolved in 100 mL of solution?

Detailed Solution: Question 18

No. of moles in 0.365 g of HCl = 0.365/36.5 = 0.01
Volume of solution in L = 100/1000 = 0.1 L
Molarity = n/V (in L) = 0.01/0.1 = 0.1 M

Test: Stoichiometry and Stoichiometric Calculations - Question 19
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What will be the molality of the solution made by dissolving 10 g of NaOH in 100 g of water?

Detailed Solution: Question 19

Molality =

Test: Stoichiometry and Stoichiometric Calculations - Question 20
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What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass?

Detailed Solution: Question 20

15 ppm = 15/106 x 102 = 1.5 x 10-3g
Molality of CHCl3 solution = 1.5 x 10-3/100 x 1000/119.5
= 1.25 x 10-4 m

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About this NEET Practice Test

This test contains 20 MCQs on Stoichiometry and Stoichiometric Calculations with detailed solutions for each question. It is part of the NCERT Based Tests for NEET course on EduRev, designed to align with the current NEET syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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