NEET Exam  >  NEET Tests  >  Chemistry Class 11  >  Test: Stoichiometry and Stoichiometric Calculations - NEET MCQ

Test: Stoichiometry and Stoichiometric Calculations - NEET MCQ


Test Description

20 Questions MCQ Test Chemistry Class 11 - Test: Stoichiometry and Stoichiometric Calculations

Test: Stoichiometry and Stoichiometric Calculations for NEET 2024 is part of Chemistry Class 11 preparation. The Test: Stoichiometry and Stoichiometric Calculations questions and answers have been prepared according to the NEET exam syllabus.The Test: Stoichiometry and Stoichiometric Calculations MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Stoichiometry and Stoichiometric Calculations below.
Solutions of Test: Stoichiometry and Stoichiometric Calculations questions in English are available as part of our Chemistry Class 11 for NEET & Test: Stoichiometry and Stoichiometric Calculations solutions in Hindi for Chemistry Class 11 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Stoichiometry and Stoichiometric Calculations | 20 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study Chemistry Class 11 for NEET Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Stoichiometry and Stoichiometric Calculations - Question 1

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

Test: Stoichiometry and Stoichiometric Calculations - Question 3

If 40 g of CaCO3 is treated with 40 g of HCl, which of the reactants will act as limiting reagent?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

How much oxygen is required for complete combustion of 560 g of ethene?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

What is the mass percent of oxygen in ethanol?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

How much mass of sodium acetate is required to make 250 mL of 0.575 molar aqueous solution?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

How much copper is present in 50 g of CuSO4?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - Question 9

Mass percent of X

Test: Stoichiometry and Stoichiometric Calculations - Question 10

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

2.82 g of glucose is dissolved in 30 g of water. The mole fraction of glucose in the solution is

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - Question 13

M1V= M2V= 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

A solution is made by dissolving 49 g of H2SO4 in 250 mL of water. The molarity of the solution prepared is

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - Question 14

Molarity

= 2 M

Test: Stoichiometry and Stoichiometric Calculations - Question 15

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

What volume of 5 M Na2SO4 must be added to 25 mL of 1 M BaCl2 to produce 10 g of BaSO4?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

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 for Test: Stoichiometry and Stoichiometric Calculations - 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

What will be the molality of the solution made by dissolving 10 g of NaOH in 100 g of water?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - Question 19

Molality =

Test: Stoichiometry and Stoichiometric Calculations - Question 20

What will be the molality of chloroform in the water sample which contains 15 ppm chloroform by mass?

Detailed Solution for Test: Stoichiometry and Stoichiometric Calculations - 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

195 videos|337 docs|190 tests
Information about Test: Stoichiometry and Stoichiometric Calculations Page
In this test you can find the Exam questions for Test: Stoichiometry and Stoichiometric Calculations solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Stoichiometry and Stoichiometric Calculations, EduRev gives you an ample number of Online tests for practice
195 videos|337 docs|190 tests
Download as PDF

How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!