Some Basic Concepts of Chemistry: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Some Basic Concepts of Chemistry  
 
(January 2026) 
 
 
Q1: For the given reaction; 
CaCO
3
 + 2HCl ? CaCl
2
 + H
2
O + CO
2 
If 90 g CaCO
3
 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 
1.13 g mL
-1
, then which of the following option is correct? 
Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol
-1
 respectively. 
A: 64.97 g of HCl remains unreacted 
B: 60.32 g of HCl remains unreacted 
C: 32.85 g of CaCO3 remains unreacted 
D: 97.30 g of HCl reacted 
Answer: A 
Explanation:  
Density of HCl solution (d) = 1.13 g/mL 
V = 300 mL 
Wt. of HCl solution = density × volume = 1.13 × 300 = 339 g 
 
CaCO3 + 2HCl ? CaCl2 + H2O + CO2 
Molar mass of CaCO3 = 40 + 12 + (3 × 16) = 100 g mol?¹ 
 
From the equation, 1 mol CaCO3 needs 2 mol HCl. 
So, 0.90 mol CaCO3 needs 2 × 0.90 = 1.80 mol HCl. 
Since available HCl = 3.58 mol and required HCl = 1.80 mol, HCl is in excess and CaCO3 is the 
limiting reagent (LR). 
Moles of HCl remained = 3.58 - 1.80 = 1.78 mol. 
Mass of HCl remained = 1.78 × 36.5 = 64.97 g. 
 
Q2: Identify the correct statements: 
A. Hydrated salts can be used as primary standard. 
B. Primary standard should not undergo any reaction with air. 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Some Basic Concepts of Chemistry  
 
(January 2026) 
 
 
Q1: For the given reaction; 
CaCO
3
 + 2HCl ? CaCl
2
 + H
2
O + CO
2 
If 90 g CaCO
3
 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 
1.13 g mL
-1
, then which of the following option is correct? 
Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol
-1
 respectively. 
A: 64.97 g of HCl remains unreacted 
B: 60.32 g of HCl remains unreacted 
C: 32.85 g of CaCO3 remains unreacted 
D: 97.30 g of HCl reacted 
Answer: A 
Explanation:  
Density of HCl solution (d) = 1.13 g/mL 
V = 300 mL 
Wt. of HCl solution = density × volume = 1.13 × 300 = 339 g 
 
CaCO3 + 2HCl ? CaCl2 + H2O + CO2 
Molar mass of CaCO3 = 40 + 12 + (3 × 16) = 100 g mol?¹ 
 
From the equation, 1 mol CaCO3 needs 2 mol HCl. 
So, 0.90 mol CaCO3 needs 2 × 0.90 = 1.80 mol HCl. 
Since available HCl = 3.58 mol and required HCl = 1.80 mol, HCl is in excess and CaCO3 is the 
limiting reagent (LR). 
Moles of HCl remained = 3.58 - 1.80 = 1.78 mol. 
Mass of HCl remained = 1.78 × 36.5 = 64.97 g. 
 
Q2: Identify the correct statements: 
A. Hydrated salts can be used as primary standard. 
B. Primary standard should not undergo any reaction with air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
D. Primary standard should not be soluble in water. 
E. Primary standard should have low relative molar mass. 
Choose the correct answer from the options given below : 
A: D and E Only 
B: A, B and E Only 
C: A, B, C and E Only 
D: A, B and C Only 
Answer: D 
Explanation: 
Primary standard (as per NCERT idea) should be a substance which is pure, stable in air, 
non-hygroscopic, readily soluble in water, and reacts completely with the titrant in a 
stoichiometric way. 
Now check each statement: 
A. Hydrated salts can be used as primary standard. 
Some hydrated salts are su?ciently stable and can be used (example often cited: Mohr’s salt). 
So this can be correct. 
B. Primary standard should not undergo any reaction with air. 
Correct. It should not react with O
2
, CO
2
 
 or moisture of air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
Correct. The reaction should be fast and have a de?nite stoichiometry so the end point is sharp. 
D. Primary standard should not be soluble in water. 
Incorrect. It should be readily soluble so that a standard solution can be prepared 
E. Primary standard should have low relative molar mass. 
Incorrect. It should preferably have high molar mass to minimise weighing error. 
So, correct statements are A, B and C only. 
 
Q3: A + 2B ? AB2 
36.0 g of 'A' (Molar mass : 60 g mol
-1
) and 56.0 g of ' B ' (Molar mass: 80 gmol
-1
) are allowed to 
react. Which of the following statements are correct ? 
A. 'A' is the limiting reagent. 
B. 77.0 g of AB2 is formed. 
C. Molar mass of AB2 is 140 g mol?¹. 
D. 15.0 g of A is left unreacted after the completion of reaction. 
Choose the correct answer from the options given below: 
A: A and C Only 
B: A and B Only 
C: C and D Only 
D: B and D Only 
Answer: D 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Some Basic Concepts of Chemistry  
 
(January 2026) 
 
 
Q1: For the given reaction; 
CaCO
3
 + 2HCl ? CaCl
2
 + H
2
O + CO
2 
If 90 g CaCO
3
 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 
1.13 g mL
-1
, then which of the following option is correct? 
Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol
-1
 respectively. 
A: 64.97 g of HCl remains unreacted 
B: 60.32 g of HCl remains unreacted 
C: 32.85 g of CaCO3 remains unreacted 
D: 97.30 g of HCl reacted 
Answer: A 
Explanation:  
Density of HCl solution (d) = 1.13 g/mL 
V = 300 mL 
Wt. of HCl solution = density × volume = 1.13 × 300 = 339 g 
 
CaCO3 + 2HCl ? CaCl2 + H2O + CO2 
Molar mass of CaCO3 = 40 + 12 + (3 × 16) = 100 g mol?¹ 
 
From the equation, 1 mol CaCO3 needs 2 mol HCl. 
So, 0.90 mol CaCO3 needs 2 × 0.90 = 1.80 mol HCl. 
Since available HCl = 3.58 mol and required HCl = 1.80 mol, HCl is in excess and CaCO3 is the 
limiting reagent (LR). 
Moles of HCl remained = 3.58 - 1.80 = 1.78 mol. 
Mass of HCl remained = 1.78 × 36.5 = 64.97 g. 
 
Q2: Identify the correct statements: 
A. Hydrated salts can be used as primary standard. 
B. Primary standard should not undergo any reaction with air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
D. Primary standard should not be soluble in water. 
E. Primary standard should have low relative molar mass. 
Choose the correct answer from the options given below : 
A: D and E Only 
B: A, B and E Only 
C: A, B, C and E Only 
D: A, B and C Only 
Answer: D 
Explanation: 
Primary standard (as per NCERT idea) should be a substance which is pure, stable in air, 
non-hygroscopic, readily soluble in water, and reacts completely with the titrant in a 
stoichiometric way. 
Now check each statement: 
A. Hydrated salts can be used as primary standard. 
Some hydrated salts are su?ciently stable and can be used (example often cited: Mohr’s salt). 
So this can be correct. 
B. Primary standard should not undergo any reaction with air. 
Correct. It should not react with O
2
, CO
2
 
 or moisture of air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
Correct. The reaction should be fast and have a de?nite stoichiometry so the end point is sharp. 
D. Primary standard should not be soluble in water. 
Incorrect. It should be readily soluble so that a standard solution can be prepared 
E. Primary standard should have low relative molar mass. 
Incorrect. It should preferably have high molar mass to minimise weighing error. 
So, correct statements are A, B and C only. 
 
Q3: A + 2B ? AB2 
36.0 g of 'A' (Molar mass : 60 g mol
-1
) and 56.0 g of ' B ' (Molar mass: 80 gmol
-1
) are allowed to 
react. Which of the following statements are correct ? 
A. 'A' is the limiting reagent. 
B. 77.0 g of AB2 is formed. 
C. Molar mass of AB2 is 140 g mol?¹. 
D. 15.0 g of A is left unreacted after the completion of reaction. 
Choose the correct answer from the options given below: 
A: A and C Only 
B: A and B Only 
C: C and D Only 
D: B and D Only 
Answer: D 
Explanation: 
Balanced reaction: 
A + 2B ? AB2 
Step 1: Calculate moles of reactants 
 
Step 2: Find the limiting reagent (NCERT method: compare required moles) 
For 0.60 mol of A, required moles of B: 
n ?(required) = 2 × 0.60 = 1.20 mol 
But available B = 0.70 mol, which is less than 1.20 mol. 
So, B is the limiting reagent. 
Hence statement A is false. 
Step 3: Moles and mass of product formed 
From the reaction, 2 mol B gives 1 mol AB2. 
So moles of AB2 formed: 
 
Molar mass of AB2: 
 
So statement C is false. 
Mass of AB2 formed: 
 
So statement B is true. 
Step 4: Unreacted A 
Moles of A consumed = moles of AB2 formed (ratio 1 : 1) 
n
A
(consumed) = 0.35 mol 
Mass of A consumed: 
m
A
(consumed) = 0.35 × 60 = 21.0 g 
Unreacted A: 
m
A
(left) = 36.0 - 21.0 = 15.0 g 
So statement D is true. 
 
Q4: In the reaction, 
2Al(s) + 6HCl(aq) ? 2Al³?(aq) + 6Cl?(aq) + 3H2(g) 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Some Basic Concepts of Chemistry  
 
(January 2026) 
 
 
Q1: For the given reaction; 
CaCO
3
 + 2HCl ? CaCl
2
 + H
2
O + CO
2 
If 90 g CaCO
3
 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 
1.13 g mL
-1
, then which of the following option is correct? 
Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol
-1
 respectively. 
A: 64.97 g of HCl remains unreacted 
B: 60.32 g of HCl remains unreacted 
C: 32.85 g of CaCO3 remains unreacted 
D: 97.30 g of HCl reacted 
Answer: A 
Explanation:  
Density of HCl solution (d) = 1.13 g/mL 
V = 300 mL 
Wt. of HCl solution = density × volume = 1.13 × 300 = 339 g 
 
CaCO3 + 2HCl ? CaCl2 + H2O + CO2 
Molar mass of CaCO3 = 40 + 12 + (3 × 16) = 100 g mol?¹ 
 
From the equation, 1 mol CaCO3 needs 2 mol HCl. 
So, 0.90 mol CaCO3 needs 2 × 0.90 = 1.80 mol HCl. 
Since available HCl = 3.58 mol and required HCl = 1.80 mol, HCl is in excess and CaCO3 is the 
limiting reagent (LR). 
Moles of HCl remained = 3.58 - 1.80 = 1.78 mol. 
Mass of HCl remained = 1.78 × 36.5 = 64.97 g. 
 
Q2: Identify the correct statements: 
A. Hydrated salts can be used as primary standard. 
B. Primary standard should not undergo any reaction with air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
D. Primary standard should not be soluble in water. 
E. Primary standard should have low relative molar mass. 
Choose the correct answer from the options given below : 
A: D and E Only 
B: A, B and E Only 
C: A, B, C and E Only 
D: A, B and C Only 
Answer: D 
Explanation: 
Primary standard (as per NCERT idea) should be a substance which is pure, stable in air, 
non-hygroscopic, readily soluble in water, and reacts completely with the titrant in a 
stoichiometric way. 
Now check each statement: 
A. Hydrated salts can be used as primary standard. 
Some hydrated salts are su?ciently stable and can be used (example often cited: Mohr’s salt). 
So this can be correct. 
B. Primary standard should not undergo any reaction with air. 
Correct. It should not react with O
2
, CO
2
 
 or moisture of air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
Correct. The reaction should be fast and have a de?nite stoichiometry so the end point is sharp. 
D. Primary standard should not be soluble in water. 
Incorrect. It should be readily soluble so that a standard solution can be prepared 
E. Primary standard should have low relative molar mass. 
Incorrect. It should preferably have high molar mass to minimise weighing error. 
So, correct statements are A, B and C only. 
 
Q3: A + 2B ? AB2 
36.0 g of 'A' (Molar mass : 60 g mol
-1
) and 56.0 g of ' B ' (Molar mass: 80 gmol
-1
) are allowed to 
react. Which of the following statements are correct ? 
A. 'A' is the limiting reagent. 
B. 77.0 g of AB2 is formed. 
C. Molar mass of AB2 is 140 g mol?¹. 
D. 15.0 g of A is left unreacted after the completion of reaction. 
Choose the correct answer from the options given below: 
A: A and C Only 
B: A and B Only 
C: C and D Only 
D: B and D Only 
Answer: D 
Explanation: 
Balanced reaction: 
A + 2B ? AB2 
Step 1: Calculate moles of reactants 
 
Step 2: Find the limiting reagent (NCERT method: compare required moles) 
For 0.60 mol of A, required moles of B: 
n ?(required) = 2 × 0.60 = 1.20 mol 
But available B = 0.70 mol, which is less than 1.20 mol. 
So, B is the limiting reagent. 
Hence statement A is false. 
Step 3: Moles and mass of product formed 
From the reaction, 2 mol B gives 1 mol AB2. 
So moles of AB2 formed: 
 
Molar mass of AB2: 
 
So statement C is false. 
Mass of AB2 formed: 
 
So statement B is true. 
Step 4: Unreacted A 
Moles of A consumed = moles of AB2 formed (ratio 1 : 1) 
n
A
(consumed) = 0.35 mol 
Mass of A consumed: 
m
A
(consumed) = 0.35 × 60 = 21.0 g 
Unreacted A: 
m
A
(left) = 36.0 - 21.0 = 15.0 g 
So statement D is true. 
 
Q4: In the reaction, 
2Al(s) + 6HCl(aq) ? 2Al³?(aq) + 6Cl?(aq) + 3H2(g) 
A. 12 L HCl(aq) is consumed for every 6 L H2(g) produced. 
B. 11.2 L H2(g) at STP is produced for every mole of HCl consumed. 
C. 33.6 L H2(g) is produced regardless of temperature and pressure for every mole of Al that 
reacts. 
D. 67.2 L H2(g) at STP is produced for every mole of Al that reacts. 
Answer: B 
Explanation: 
2Al(s) + 6HCl(aq) ? 2Al³?(aq) + 6Cl?(aq) + 3H2(g) 
From the balanced equation, 6 moles of HCl give 3 moles of H2. 
So, moles of H2 produced  
 
 
Q5: By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The 
percentage of phosphorus in compound (X) is : (nearest integer) 
(Given, molar mass in g mol-1: O = 16, Mg = 24, P = 31) 
A: 40 
B: 30 
C: 20 
D: 50 
Answer: D 
Explanation: 
The chemical formula for magnesium pyrophosphate is Mg2P2O7. 
Using the given molar masses (Mg = 24, P = 31, O = 16): 
Molar mass of Mg2P2O7 = (2 × 24) + (2 × 31) + (7 × 16) 
Molar mass = 48 + 62 + 112 = 222 g mol?¹ 
One mole of Mg2P2O7 contains 2 moles of phosphorus atoms. 
Mass of P = 2 × 31 = 62 g 
This means that 222 g of Mg2P2O7 contains 62 g of phosphorus. 
The mass of Mg2P2O7 obtained in the experiment is 1.79 g. 
Applying the unitary method: 
Mass of P in 1.79 g of Mg2P2O7 = (62/222) × 1.79 g 
Mass of P = 110.98 / 222 ˜ 0.4999 g 
The percentage of an element in a given compound is calculated as: 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Some Basic Concepts of Chemistry  
 
(January 2026) 
 
 
Q1: For the given reaction; 
CaCO
3
 + 2HCl ? CaCl
2
 + H
2
O + CO
2 
If 90 g CaCO
3
 is added to 300 mL of HCl which contains 38.55% HCl by mass and has density 
1.13 g mL
-1
, then which of the following option is correct? 
Given molar mass of H, Cl, Ca and O are 1, 35.5, 40 and 16 g mol
-1
 respectively. 
A: 64.97 g of HCl remains unreacted 
B: 60.32 g of HCl remains unreacted 
C: 32.85 g of CaCO3 remains unreacted 
D: 97.30 g of HCl reacted 
Answer: A 
Explanation:  
Density of HCl solution (d) = 1.13 g/mL 
V = 300 mL 
Wt. of HCl solution = density × volume = 1.13 × 300 = 339 g 
 
CaCO3 + 2HCl ? CaCl2 + H2O + CO2 
Molar mass of CaCO3 = 40 + 12 + (3 × 16) = 100 g mol?¹ 
 
From the equation, 1 mol CaCO3 needs 2 mol HCl. 
So, 0.90 mol CaCO3 needs 2 × 0.90 = 1.80 mol HCl. 
Since available HCl = 3.58 mol and required HCl = 1.80 mol, HCl is in excess and CaCO3 is the 
limiting reagent (LR). 
Moles of HCl remained = 3.58 - 1.80 = 1.78 mol. 
Mass of HCl remained = 1.78 × 36.5 = 64.97 g. 
 
Q2: Identify the correct statements: 
A. Hydrated salts can be used as primary standard. 
B. Primary standard should not undergo any reaction with air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
D. Primary standard should not be soluble in water. 
E. Primary standard should have low relative molar mass. 
Choose the correct answer from the options given below : 
A: D and E Only 
B: A, B and E Only 
C: A, B, C and E Only 
D: A, B and C Only 
Answer: D 
Explanation: 
Primary standard (as per NCERT idea) should be a substance which is pure, stable in air, 
non-hygroscopic, readily soluble in water, and reacts completely with the titrant in a 
stoichiometric way. 
Now check each statement: 
A. Hydrated salts can be used as primary standard. 
Some hydrated salts are su?ciently stable and can be used (example often cited: Mohr’s salt). 
So this can be correct. 
B. Primary standard should not undergo any reaction with air. 
Correct. It should not react with O
2
, CO
2
 
 or moisture of air. 
C. Reactions of primary standard with another substance should be instantaneous and 
stoichiometric. 
Correct. The reaction should be fast and have a de?nite stoichiometry so the end point is sharp. 
D. Primary standard should not be soluble in water. 
Incorrect. It should be readily soluble so that a standard solution can be prepared 
E. Primary standard should have low relative molar mass. 
Incorrect. It should preferably have high molar mass to minimise weighing error. 
So, correct statements are A, B and C only. 
 
Q3: A + 2B ? AB2 
36.0 g of 'A' (Molar mass : 60 g mol
-1
) and 56.0 g of ' B ' (Molar mass: 80 gmol
-1
) are allowed to 
react. Which of the following statements are correct ? 
A. 'A' is the limiting reagent. 
B. 77.0 g of AB2 is formed. 
C. Molar mass of AB2 is 140 g mol?¹. 
D. 15.0 g of A is left unreacted after the completion of reaction. 
Choose the correct answer from the options given below: 
A: A and C Only 
B: A and B Only 
C: C and D Only 
D: B and D Only 
Answer: D 
Explanation: 
Balanced reaction: 
A + 2B ? AB2 
Step 1: Calculate moles of reactants 
 
Step 2: Find the limiting reagent (NCERT method: compare required moles) 
For 0.60 mol of A, required moles of B: 
n ?(required) = 2 × 0.60 = 1.20 mol 
But available B = 0.70 mol, which is less than 1.20 mol. 
So, B is the limiting reagent. 
Hence statement A is false. 
Step 3: Moles and mass of product formed 
From the reaction, 2 mol B gives 1 mol AB2. 
So moles of AB2 formed: 
 
Molar mass of AB2: 
 
So statement C is false. 
Mass of AB2 formed: 
 
So statement B is true. 
Step 4: Unreacted A 
Moles of A consumed = moles of AB2 formed (ratio 1 : 1) 
n
A
(consumed) = 0.35 mol 
Mass of A consumed: 
m
A
(consumed) = 0.35 × 60 = 21.0 g 
Unreacted A: 
m
A
(left) = 36.0 - 21.0 = 15.0 g 
So statement D is true. 
 
Q4: In the reaction, 
2Al(s) + 6HCl(aq) ? 2Al³?(aq) + 6Cl?(aq) + 3H2(g) 
A. 12 L HCl(aq) is consumed for every 6 L H2(g) produced. 
B. 11.2 L H2(g) at STP is produced for every mole of HCl consumed. 
C. 33.6 L H2(g) is produced regardless of temperature and pressure for every mole of Al that 
reacts. 
D. 67.2 L H2(g) at STP is produced for every mole of Al that reacts. 
Answer: B 
Explanation: 
2Al(s) + 6HCl(aq) ? 2Al³?(aq) + 6Cl?(aq) + 3H2(g) 
From the balanced equation, 6 moles of HCl give 3 moles of H2. 
So, moles of H2 produced  
 
 
Q5: By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The 
percentage of phosphorus in compound (X) is : (nearest integer) 
(Given, molar mass in g mol-1: O = 16, Mg = 24, P = 31) 
A: 40 
B: 30 
C: 20 
D: 50 
Answer: D 
Explanation: 
The chemical formula for magnesium pyrophosphate is Mg2P2O7. 
Using the given molar masses (Mg = 24, P = 31, O = 16): 
Molar mass of Mg2P2O7 = (2 × 24) + (2 × 31) + (7 × 16) 
Molar mass = 48 + 62 + 112 = 222 g mol?¹ 
One mole of Mg2P2O7 contains 2 moles of phosphorus atoms. 
Mass of P = 2 × 31 = 62 g 
This means that 222 g of Mg2P2O7 contains 62 g of phosphorus. 
The mass of Mg2P2O7 obtained in the experiment is 1.79 g. 
Applying the unitary method: 
Mass of P in 1.79 g of Mg2P2O7 = (62/222) × 1.79 g 
Mass of P = 110.98 / 222 ˜ 0.4999 g 
The percentage of an element in a given compound is calculated as: 
 
Rounding off to the nearest integer, we get 50%. 
Option D is correct. 
 
Q6: Aqueous HCl reacts with MnO
2
(s) to form MnCl
2
(aq), Cl
2
(g), and H
2
O(l). What is the weight 
(in g) of Cl
2
 liberated when 8.7 g of MnO
2
(s) is reacted with excess aqueous HCl solution? 
(Given Molar mass in g mol-1 Mn = 55, Cl = 35.5, O = 16, H = 1) 
A: 14.2 
B: 21.3  
C: 7.1 
D: 71 
Answer: C 
Explanation: 
Balance the reaction ?rst (NCERT method): 
MnO2(s) + 4HCl(aq) ? MnCl2(aq) + Cl2(g) + 2H2O(l) 
Step 1: Calculate moles of MnO2 
Molar mass of MnO2: 
M(MnO2) = 55 + 2(16) = 55 + 32 = 87 g mol?¹ 
Moles of MnO2 in 8.7 g: 
n(MnO2) = 8.7 / 87 = 0.1 mol 
Step 2: Use mole ratio to ?nd moles of Cl2 
From the balanced equation: 
1 mol MnO2 ? 1 mol Cl2 
So, 
n(Cl2) = 0.1 mol 
Step 3: Convert moles of Cl2 to mass 
Molar mass of Cl2 = 2 × 35.5 = 71 g mol?¹ 
Mass of Cl2 = 0.1 × 71 = 7.1 g 
Correct option: C (7.1 g) 
 
Q7: 14.0 g of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273 K. 
Which of the following statements is incorrect? 
[Given: Molar mass in g mol?¹ of Ca = 40, Cl = 35.5, H = 1] 
A: The limiting reagent is calcium metal. 
B: 33.3 g of CaCl2 is produced. 
C: 7.84 L of H2 gas is evolved. 
D: 0.35 mol of H2 gas is evolved. 
Answer: B