NCERT Solutions Class 11 Chemistry - Basic Concepts of Chemistry

Q1.1: Calculate the molecular mass of the following:
(i) H₂O
(ii) CO₂
(iii) CH₄
Ans: (i) H₂O

H₂O Structure

The molecular mass of water, H₂O:
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084)  + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u
(ii) CO₂

 CO₂ Structure

The molecular mass of carbon dioxide, CO_2:
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u +  32.00 u
= 44.01 u
(iii) CH₄

The molecular mass of methane, CH_4:
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u  
= 16.043 u

Q1.2: Calculate the mass per cent of different elements present in sodium sulfate (Na₂SO₄).
Ans:

Na₂SO₄ Structure

The molecular formula of sodium sulphate is Na₂SO₄
Molar mass of Na₂SO₄ = [(2 × 23.0) + (32.066) + 4 (16.00)]
= 142.066 g
Mass percent of an element:

Mass percent of sodium:

= 32.379
= 32.4%
Mass percent of sulphur:

= 22.57
= 22.6 %
Mass percent of oxygen:

= 45.049 = 45.05%

Q1.3: Determine the empirical formula of an oxide of iron that has 69.9% iron & 30.1% dioxygen by mass.
Ans:

Given:
% of iron by mass = 69.9 %
% of oxygen by mass = 30.1 %
Relative moles of iron in iron oxide:

 = 1.25

Relative moles of oxygen in iron oxide:

 = 1.88

The simplest molar ratio of iron to oxygen = 1.25: 1.88 = 1:1.5 = 2:3
∴ The empirical formula of the iron oxide is Fe₂O₃.

Q1.4: Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in the air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Ans:
The balanced reaction of combustion of carbon can be written as:

(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide, i.e 44g of CO₂.
(ii) According to the question, only 16 g (0.5 mol) of dioxygen is available. Hence, it will react with 0.5 moles of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.
(iii) Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O₂ will react with 6 g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18 g of carbon (1.5 mol) will not undergo combustion.

Q1.5: Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molars aqueous solution. The molar mass of sodium acetate is 82.0245 g mol^–1.
Ans: 0.375 M aqueous solution of sodium acetate ≡ 1000 mL of a solution containing 0.375 moles of sodium acetate.
∴ Number of moles of sodium acetate in 500 mL.
= 0.375/1000 x 500
= 0.1875 mole
Molar mass of sodium acetate = 82.0245 g mol^–1 (Given)
∴ Required mass of sodium acetate = (82.0245 g mol^–1) (0.1875 mole)
= 15.38 g

Q1.6: Calculate the concentration of nitric acid in moles per litre in a sample that has a density, 1.41 g mL^–1, and the mass percent of nitric acid in it being 69%.
Ans: The mass percent of nitric acid in the sample = 69 % (Given)
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO₃):
= [1 +14 + 3(16)] g mol^–1
= 1 +14 + 48
= 63 g mol^–1
∴ Number of moles in 69 g of HNO₃
= 1.095 mol
The volume of 100 g of nitric acid solution:

= 70.92 mL ≡ 70.92 x 10^-3 L
The concentration of nitric acid:

= 15.44 mol/L
∴ Concentration of nitric acid = 15.44 mol/L

Q1.7: How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Ans: 1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO₄ = (63.5) + (32.00) + 4(16.00)
= 63.5 +32.00 + 64.00
= 159.5 g
159.5 g of CuSO₄ contains 63.5 g of copper.
⇒ 100 g of CuSO₄ will contain  
 of copper.
∴ Amount of copper that can be obtained from 100 g CuSO₄ 


Q1.8: Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. 
Ans: Mass percent of iron (Fe) = 69.9% (Given)
Mass percent of oxygen (O) = 30.1% (Given)
Number of moles of iron present in the oxide = 69.90/55.85 = 1.25
Number of moles of oxygen present in the oxide = 30.1/16.0 = 1.88
The ratio of iron to oxygen in the oxide:

The empirical formula of the oxide is Fe₂O_3.
Empirical formula mass of Fe₂O₃ = [2(55.85) + 3(16.00)] g
Molar mass of Fe₂O₃ = 159.69 g

The molecular formula of a compound is obtained by multiplying the empirical formula with n.
Thus, the empirical formula of the given oxide is Fe₂O₃ and n is 1.
Hence, the molecular formula of the oxide is Fe₂O_3.

Q1.9: Calculate the atomic mass (average) of chlorine using the following data:

Ans: Fractional Abundance of ³⁵Cl = 0.7577 and Molar mass = 34.9689  
Fractional Abundance of ³⁷Cl = 0.2423 and Molar mass = 36.9659
Average Atomic mass = (0.7577 x 34.9689)amu + (0.2423 x 36.9659)
= 26.4959 + 8.9568 = 35.4527

Q1.10: In three moles of ethane (C₂H₆), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Ans: (i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.
Number of moles of carbon atoms in 3 moles of C₂H₆ 
= 2 × 3 = 6
(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
Number of moles of carbon atoms in 3 moles of C₂H₆ 
= 3 × 6 = 18
(iii) 1 mole of C₂H₆ contains 6.023 × 10²³ molecules of ethane.
Number of molecules in 3 moles of C₂H₆ 
= 3 × 6.023 × 10²³
= 18.069 × 10²³

Q1.11: What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^–1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?+
Ans: Molarity (M) of a solution is given by:


The molar concentration of sugar = 0.02925 mol L^–1

Q1.12: If the density of methanol is 0.793 kg L^–1, what is the volume needed for making 2.5 L of its 0.25 M solution?
Ans:
CH₃OH
Molar mass of methanol (CH₃OH) = (1 × 12) + (4 × 1) + (1 × 16)
= 32 g mol^–1
= 0.032 kg mol^–1
Molarity of methanol solution:

(Since density is mass per unit volume)
Applying,
M₁V₁ = M₂V₂
(Given solution) = (Solution to be prepared)
(24.78 mol L^–1) V₁ = (2.5 L) (0.25 mol L^–1)
V₁ = 0.0252 L
V₁ = 25.22 mL

Q1.13: The pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below:
1 Pa = 1 N m^–2
If the mass of air at sea level is 1034 g cm^–2, calculate the pressure in Pascals.
Ans: The pressure is defined as a force acting per unit area of the surface.
P = F/A

= 1.01332 × 10⁵ kg m^-1 s^-2
We know,
1 N = 1 kg ms^–2
Then,
1 Pa = 1 N m^–2 = 1 kg m^–2s^–2
1 Pa = 1 kg m^1–s^2–
∴ Pressure = 1.01332 × 10⁵ Pa.

Q1.14: What is the SI unit of mass? How is it defined?
Ans: The SI unit of mass is the kilogram (kg).
1 Kilogram is defined as the mass equal to the mass of the international prototype of the kilogram.

Q1.15: Match the following prefixes with their multiples

Ans:

Q1.16: What do you mean by significant figures?
Ans: Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value.
Example: If 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures is 3.
Hence, significant figures are defined as the total number of digits in a number, including the last digit that represents the uncertainty of the result.

Q1.17: A sample of drinking water was found to be severely contaminated with chloroform, CHCl₃, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample.
Ans: (i) 1 ppm is equivalent to 1 part out of 1 million (10⁶) parts.
The mass percent of 15 ppm chloroform in water:
= 15/10⁶ × 100
≈1.5 x 10^-3%
(ii) 100 g of the sample contains 1.5 × 10^–3 g of CHCl_3.
⇒ 1000 g of the sample contains 1.5 × 10^–2 g of CHCl_3.
The molality of chloroform in water:

Molar mass of CHCl₃ = 12.00 + 1.00 + 3(35.5)
= 119.5 g mol^–1
Molality of chloroform in water = 0.0125 × 10^–2 m
= 1.25 × 10^–4 m

Q1.18: Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Ans: (i) 0.0048 = 4.8× 10^–3
(ii) 234, 000 = 2.34 × 10⁵
(iii) 8008 = 8.008 × 10³
(iv) 500.0 = 5.000 × 10²
(v) 6.0012 = 6.0012

Q1.19: How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Ans: (i) 0.0025
There are 2 significant figures.
(ii) 208
There are 3 significant figures.
(iii) 5005
There are 4 significant figures.
(iv) 126,000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi) 2.0034
There are 5 significant figures.

Q1.20: Round up the following up to three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Ans: (i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810

Q1.21: The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ........... mm = ........... pm
(ii) 1 mg = ...........  kg = ...........  ng
(iii) 1 ml = ...........  l = ...........  dm³
Ans: (a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g.
The masses of dioxygen bear a whole number ratio of 1:2:1:5. Hence, the given experimental data obeys the law of multiple proportions.
The law states that if two elements combine to form more than one compound, then the masses of one element that combine with the fixed mass of another element are in the ratio of small whole numbers.
(b) (i) 1 km = 1 km × 1000 m/1 km × 100 cm/1m × 10 mm/1cm
⇒ 1 km = 10⁶ mm
⇒ 1 km = 1 km × 1000 m/1 km × 1 pm/10^-12 m
⇒ 1 km = 10¹⁵ pm
Hence, 1 km = 10⁶ mm = 10¹⁵ pm
(ii) 1 mg = 1 mg × 1 g/1000 mg × 1 kg/1000 g
⇒ 1 mg = 10^-6 kg
⇒ 1 mg = 1 mg × 1 g/ 1000 mg × 1 mg/ 10^-9 g 1
⇒ 1 mg = 10⁶ ng
Hence, 1 mg = 10^–6 kg = 10⁶ ng
(iii) 1 mL = 1 mL × 1 L/1000 mL
⇒ 1 mL = 10^–3 L
⇒ 1 mL = 1 cm³ 


Q1.22: If the speed of light is 3.0 × 10⁸ ms^-1, calculate the distance covered by light in 2.00 ns.
Ans: According to the question:
Time taken to cover the distance = 2.00 ns = 2.00 × 10^–9 s
Speed of light = 3.0 × 10⁸ ms^–1
Distance travelled by light in 2.00 ns:
= Speed of light × Time taken
= (3.0 × 10⁸ ms ^–1) (2.00 × 10^–9 s)
= 6.00 × 10^–1 m
= 0.600 m

Q1.23: In a reaction 
A + B₂ → AB₂
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B₂ 
(ii) 2 mol A + 3 mol B₂
(iii) 100 atoms of A + 100 molecules of B₂ 
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Ans: A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

Q1.24: Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N₂(g) + H₂(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00 × 10³ g of dinitrogen reacts with 1.00 × 10³ g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Ans:
(i) Balancing the given chemical equation,
 N_2(g) + H_2(g) → 2NH_3(g)
From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole (34 g) of ammonia.
⇒ 2.00 × 10³ g of dinitrogen will react with 6 g/28 g × 2.00 x 10³ g dihydrogen i.e., 2.00 × 10³ g of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 10³ g
Hence, N₂ is the limiting reagent. 28 g of N₂ produces 34 g of NH_3.
Hence, mass of ammonia produced by:
2000 g of N₂ = 34 g/28 g × 2000g
= 2428.57 g
(b) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.
(c) Mass of dihydrogen left unreacted = 1.00 × 10³ g – 428.6 g = 571.4 g

Q1.25: How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Ans:

Molar mass of Na₂CO₃ = (2 × 23) +12.00 + (3 × 16) = 106 g mol^–1
Now, 1 mole of Na₂CO₃ means 106 g of Na₂CO_3.
⇒ 0.5 mol of Na₂CO₃ 

= 53 g Na₂CO₃
⇒ 0.50 M of Na₂CO₃ = 0.50 mol/L Na₂CO₃
Hence, 0.50 mol of Na₂CO₃ is present in 1 L of water or 53 g of Na₂CO₃ is present in 1 L of water.

Q1.26: If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Ans: Reaction of dihydrogen with dioxygen can be written as:
2H_2(g) + O_2(g) → 2H₂O_(g)
Now, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water vapour.
Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

Q1.27: Convert the following into basic units:
(i) 28.7 pm                
(ii) 15.15 pm                    
(iii) 25365 mg
Ans: (i) 28.7 pm
1 pm = 10^–12 m
28.7 pm = 28.7 × 10^–12 m
= 2.87 × 10^–11 m
(ii) 15.15 pm
1 pm = 10^–12 m
15.15 pm = 15.15 × 10^–12 m
= 1.515 × 10^–11 m
(iii) 25365 mg
1 mg = 10^–3 g
25365 mg = 2.5365 × 10⁴ × 10^–3 g
Since,
1 g = 10^–3 kg
2.5365 × 10¹ g = 2.5365 × 10^–1 × 10^–3 kg
25365 mg = 2.5365 × 10^–2 kg.

Q1.28: Which one of the following will have largest number of atoms?
(i) 1 g Au (s)  
(ii) 1 g Na (s)
(iii) 1 g Li (s)  
(iv) 1 g of Cl₂(g)
Ans: (i) 1 g of Au (s)
= 1/197 mol of Au (s) = 6.022 × 10²³/197
atoms of Au (s) = 3.06 × 10²¹ atoms of Au (s)
(ii) 1 g of Na (s) = ¹/₂₃ mol of Na (s)
= 6.022 × 10²³/23  atoms of Na (s)
= 0.262 × 10²³ atoms of Na (s)
= 26.2 × 10²¹ atoms of Na (s)
(iii) 1 g of Li (s) = ¹/₇ mol of Li (s)
= 6.022 × 10²³/7 atoms of Li (s)
= 0.86 × 10²³ atoms of Li (s)
= 86.0 × 10²¹ atoms of Li (s)
(iv) 1 g of Cl₂ (g) = 1/71  mol of Cl₂ (g)
(Molar mass of Cl₂ molecule = 35.5 × 2 = 71 g mol^–1)
= 6.022 x 10²³/71
= 0.0848 × 10²³ molecules of Cl₂ (g)
= 8.48 × 10²¹ molecules of Cl₂ (g)
As one molecule of Cl₂ contains two atoms of Cl.
Number of atoms of Cl = 2 × 8.48 × 10²¹ = 16.96 × 10²¹  atoms of Cl
Hence, 1 g of Li (s) will have the largest number of atoms.

Q1.29: Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
Ans: Mole fraction of C₂H₅OH


Number of moles present in 1 L water:

Substituting the value of ⁿH₂O in equation (1)

Q1.30: What will be the mass of one 12C atom in g?
Ans: 1 mole of carbon atoms = 6.023 × 10²³ atoms of carbon
= 12 g of carbon
Mass of one ¹²C atom = 12 g/6.022 × 10²³ 
= 1.993 × 10^–23 g.

Q1.31: How many significant figures should be present in the of the following calculations?
(i)
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Ans:
(i)
Least precise number of calculation = 0.112
Number of significant figures in the Ans = Number of significant figures in the least precise number = 3
(ii) 5 × 5.364
Least precise number of calculation = 5.364
Number of significant figures in the Ans = Number of significant figures in 5.364 = 4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term is four, the number of significant figures in the Ans: is also 4.

Q1.32: Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

Ans: Molar mass of argon:

= 39.947 g mol^–1

Q1.33: Calculate the number of atoms in each of the following: 
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He
Ans: (i) 1 mole of Ar = 6.022 × 10²³ atoms of Ar
52 mol of Ar = 52 × 6.022 × 10²³ atoms of Ar
= 3.131 × 10²⁵ atoms of Ar
(ii) 4 u of He = 1 atom of He
1 u of He= ¹/₄ atom of He
52 u of He = 52 /4 atoms of He = 13 atoms of He
(iii) 4 g of He = 6.022 × 10²³ atoms of He
 = 7.8286 × 10²⁴ atoms of He.

Q1.34: A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. 
Calculate: 
(i) Empirical formula
(ii) Molar mass of the gas
(iii) Molecular formula
Ans: (i) 1 mole (44 g) of CO₂ contains 12 g of carbon.
3.38 g of CO₂ will contain carbon = 12 g/44 g × 3.38 g = 0.9217 g
18 g of water contains 2 g of hydrogen. = 2 g/18 g × 0.690
0.690 g of water will contain hydrogen = 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is: = 0.9217 g + 0.0767 g = 0.9984 g
Percent of C in the compound = 0.9217 g/0.9984 × 100 = 92.32%
Percent of H in the compound = 0.0767 g/0.9984 × 100 = 7.68%
Moles of carbon in the compound = 92.32/12.00 = 7.69
Moles of hydrogen in the compound = 7.68
∴ Ratio of carbon to hydrogen in the compound = 7.69 : 7.68 = 1 : 1
Hence, the empirical formula of the gas is CH.

(ii) Given,
Weight of 10.0 L of the gas (at S.T.P) = 11.6 g
∴ Weight of 22.4 L of gas at STP
= 25.984 g≈ 26 g.
Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g
= 26 g/13 gn = 2
Molecular formula of gas = (CH) = C₂H₂

Q1.35: Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction, 
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)
What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Ans: 0.75 M of HCl
≡ 0.75 mol of HCl are present in 1 L of water
≡ [(0.75 mol) × (36.5 g mol^–1)] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
Amount of HCl present in 25 mL of solution = 27.375 g/1000 mL × 25 mL = 0.6844 g
From the given chemical equation,
CaCO_3(g) + 2HCl_(aq) → CaCl_2(aq) + CO_2(g) + H₂O(l)
2 mol of HCl (2 × 36.5 = 73 g) react with 1 mol of CaCO₃ (100 g).
∴ Amount of CaCO₃ that will react with 0.6844 g = (100/73) × 0.6844 = 0.9375 g

Q1.36: Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4 HCl_(aq) + MnO_2(s) → 2 H₂O_(l) + MnCl_2(aq) + Cl_2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Ans: 1 mol [55 + 2 × 16 = 87 g] MnO₂ reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl.
5.0 g of MnO₂ will react with
= 146 g/87 g × 5.0 g of HCl
= 8.4 g of HCl
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.