NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry is prepared by our senior and renown teachers of Physics Wallah primary focus while solving these questions of class11 in NCERT textbook, also do read theory of this Chapter 1 Some Basic Concepts of Chemistry while going before solving the NCERT questions. Our Physics Wallah team Prepared Other Subjects NCERT Solutions for class 11.
1. Calculate the molecular mass of the following :
(i) H_{2}O
(ii) CO_{2}
(iii) CH_{4}
Solution :
(i)CH_{4} :
Molecular weight of methane, CH_{4}
= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)
= [1(12.011 u) +4 (1.008u)]
= 12.011u + 4.032 u
= 16.043 u
(ii) H_{2}O :
Molecular weight of water, H_{2}O
= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u +16.00 u
= 18.016u
So approximately
= 18.02 u
(iii) CO_{2} :
= Molecular weight of carbon dioxide, CO_{2}
= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)
= [1(12.011 u) + 2(16.00 u)]
= 12.011 u +32.00 u
= 44.011 u
So approximately
= 44.01u
2. Calculate the mass percent of different elements present in Sodium Sulphate (Na_{2}SO_{4}).
Solution :
Molar mass of Na_{2}SO_{4 }= [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g
Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100
∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%
Mass percent of sulphur(S): (32/142) × 100 = 22.54%
Mass percent of oxygen:(O): (64/142) × 100 = 45.07%
3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Solution :
% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]
Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide is Fe_{2}O_{3}.
4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in 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.
Solution :
The balanced reaction of combustion of carbon in dioxygens is:
C(s) + O2(g) → CO_{2} (g)
1mole 1mole(32g) 1mole(44g)
(i) In dioxygen, combustion is complete. Therefore 1 mole of carbon dioxide produced by burning 1 mole of carbon.
(ii) Here, oxgen acts as a limiting reagent as only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide.
(iii) Here again oxgen acts as a limiting reagent as only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.
5. Calculate the mass of sodium acetate (CH_{3}COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245g mol^{1}.
Solution :
0.375 M aqueous solution of sodium acetate means that 1000 mL of solution containing 0.375 moles of sodium acetate.
∴No. of moles of sodium acetate in 500 mL = (0.375/1000)×500 = 0.375/2 = 0.1875
Molar mass of sodium acetate = 82.0245g mol^{1}
∴Mass of sodium acetate acquired = 0.1875×82.0245 g = 15.380g
6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL1 and the mass per cent of nitric acid in it being 69%.
Solution :
Mass percent of 69% means tat 100g of nitric acid solution contain 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO3) = 1+14+48 = 63g mol^{1}
Number of moles in 69 g of HNO3 = 69/63 moles = 1.095 moles
Volume of 100g nitric acid solution = 100/1.41 mL = 70.92 mL = 0.07092 L
∴ Conc. of HNO_{3} in moles per litre = 1.095/0.07092 = 15.44 M
7. How much copper can be obtained from 100 g of copper sulphate (CuSO_{4} )?
Solution :
1 mole of CuSO_{4} contains 1 mole of copper.
Molar mass of CuSO_{4} = (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO_{4} contains 63.5 g of copper.
∴ copper can be obtained from 100 g of copper sulphate = (63.5/159.5)×100 = 39.81g
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.Given that the molar mass of the oxide is 159.69 g mol^{1}
Solution :
% of iron by mass = 69.9 % [Given]
% of oxygen by mass = 30.1 % [Given]
Atomic mass of iron = 55.85 amu
Atomic mass of oxygen = 16.00 amu
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass /Atomic mass of oxygen = 30.01/16 =1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide is Fe_{2}O_{3}.
Mass of Fe_{2}O_{3} = (2×55.85) + (3×16.00) = 159.7 g mol^{1}
n = Molar mass /Empirical formula mass = 159.7/ 159.6 = 1(approx)
Thus, Molecular formula is same as Empirical Formula i.e. Fe_{2}O_{3}.
9. Calculate the atomic mass (average) of chlorine using the following data :
% Natural Abundance Molar Mass
35Cl 75.77 34.9689
37Cl 24.23 36.9659
Solution :
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 × 34.9689) amu + (0.2423 × 36.9659)
= 26.4959 + 8.9568 = 35.4527
10. In three moles of ethane (C_{2}H_{6}), calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution :
(i) 1 mole of C_{2}H_{6 } contains 2 moles of Carbon atoms
∴ 3 moles of of C_{2}H_{6} will contain 6 moles of Carbon atoms
(ii) 1 mole of C_{2}H_{6 } contains 6 moles of Hydrogen atoms
∴ 3 moles of of C_{2}H_{6} will contain 18 moles of Hydrogen atoms
(iii) 1 mole of C_{2}H_{6} contains Avogadro's no. 6.02 ×10^{23} molecules
∴ 3 moles of of C_{2}H_{6} will contain ethane molecule = 3×6.02 × 10^{23}= 18.06 ×10^{23} molecules.
11. What is the concentration of sugar (C_{12}H_{22}O_{11}) in mol L^{1} if its 20 g are dissolved in
enough water to make a final volume up to 2L?
Solution :
Molar mass of sugar (C_{12}H_{22}O_{11}) = (12 ×12) +(1 ×22)+ (11×16) = 342 g mol^{1}
No. of moles in 20g of sugar = 20/342 = 0.0585 mole
Volume of Solution = 2L (given)
Molar concentration = Moles of solute/Volume of solution in L = 0.0585mol /2L = 0.0293 mol L1 = 0.0293 M
12. If the density of methanol is 0.793 kg L1 , what is its volume needed for making 2.5 L of its 0.25 M solution?
Solution :
Molar mass of methanol (CH_{3}OH) = (1×12) + (4×1) + (1×16) = 32 g mol1 = 0.032 kg mol^{1}
Molarity of the solution = 0.793 /0.032 = 24.78 mol L^{1 }
Applying, M_{1}V_{1 }(Given Solution) = M_{2}V_{2} (Solution to be prepared)
24.78×V_{1} = 0.25×2.5 L
V_{1}= 0.02522 L = 25.22 mL
13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below :
1Pa = 1N m^{2}
If mass of air at sea level is 1034 g cm^{2},calculate the pressure in pascal.
Solution :
Pressure is the force (i.e. weigh) acting per unit area.
P= F/A = 1034g × 9.8ms ^{2}/cm^{2}
= 1034g × 9.8ms^{2 }/cm^{2 }× 1kg /1000g × 100cm /1m × 100cm /1m = 1.01332 ×105 N
Now,
1Pa = 1N m^{2}
∴ 1.01332 × 10^{5} N ×m^{2} = 1.01332 ×10^{5} Pa
14. What is the SI unit of mass? How is it defined?
Solution :
The SI unit of mass is kilogram (kg).
The kg is defined as the mass of platinumiridium (PtIr) cylinder that is stored in an airtight jar at International Bureau of Weigh and Measures in France.
15. Match the following prefixes with their multiples:
Prefixes 
Multiples 

(a) 
femto 
10 
(b) 
giga 
10^{−15} 
(c) 
mega 
10^{−6} 
(d) 
deca 
10^{9} 
(e) 
micro 
10^{6} 
Solution :
Prefixes 
Multiples 

(a) 
femto 
10^{−15} 
(b) 
giga 
10^{9} 
(c) 
mega 
10^{6} 
(d) 
deca 
10 
(e) 
micro 
10^{−6} 
16. What do you mean by significant figures ?
Solution :
Significant figures are meaningful digits which are known with certainty including the last digit whose value is uncertain.
For example,
In 11.2546 g, there are 6 significant figures but here 11.254 is certain and 6 is uncertain and the uncertainty would be ±1 in the last digit. Hence last uncertain digit is also included in Significant figures.
17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl_{3}, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution :
(i) 15 ppm means 5 parts in million(10^{6}) parts.
∴ % by mass = 15/10^{6} × 100 = 15 × 10^{4 }= 1.5×10^{3} %
(ii) Molar mass of chloroform(CHCl^{3}) = 12+1+ (3×35.5) = 118.5 g mol^{1}
100g of the sample contain chloroform = 1.5×103g
∴ 1000 g(1 kg) of the sample will contain chloroform = 1.5×10^{2} g
= 1.5×10^{2}/ 118.65 mole = 1.266 ×10^{4} mole
∴ Molality = 1.266×10^{4} m.
18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution :
(i) 0.0048 = 4.8× 10^{3}
(ii) 234, 000 = 2.34× 10^{5}
(iii) 8008 = 8.008× 10^{3}
(iv) 500.0 = 5.000× 10^{2}
(v) 6.0012 = 6.0012× 10^{0}
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
Solution :
(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5
20. Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution :
(i) 34.2
(ii) 10.4
(iii) 0.046
(iv) 2810
21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
Mass of dinitrogen Mass of dioxygen
(i)14 g 16 g
(ii)14 g 32 g
(iii)28 g 32 g
(iv)28 g 80 g
(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 = ...................... dm^{3}
Solution :
(a) Fixing the mass of dinitrogen as 28 g, masses of dioxygen combined will be 32, 64, 32 and 80 g in the given four oxides. These masses of dioxygen bears a simple whole number ratio as 2:4:2:5. Hence, the data given will obey the law of multiple proportions.
The statement is as follows two elements always combine in a fixed mass of other bearing a simple ratio to another to form two or more chemical compounds.
(b) (i) 1 km = 1km× 1000m/ 1km ×100cm /1m/ 10mm /1cm = 10^{6} mm
1 km = 1km× 1000m / 1km × 1pm/ 10^{12}m = 10^{15 }pm
(ii) 1 mg = 1mg ×1g/ 1000mg × 1kg / 1000g = 106 kg
1 mg = 1mg ×1g/ 1000mg × 1ng/ 10^{9}g = 10^{6} ng
(iii) 1 mL = 1mL×1L/ 1000mL = 10^{3} L
1 mL = 1cm^{3} = 1cm^{3}× (1dm × 1dm × 1dm/ 10cm × 10cm × 10cm) = 10^{3}dm^{3}
22. If the speed of light is 3.0 × 108ms1, calculate the distance covered by light in 2.00 ns.
Solution :
Distance covered = Speed × Time = 3.0 × 108ms^{1} × 2.00 ns
= 3.0 × 108ms^{1} × 2.00 ns ×10^{9}s /1ns = 6.00×10^{1}m = 0.600m
23. In a reaction
A + B_{2} → AB_{2}
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Solution :
(i) According to the reaction, 1 atom of A reacts with 1 molecule of B.
∴200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unreacted. Hence, B is the limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B.
∴ 2 mol of A will react with only 2 mol of B leaving 1 mol of B. Hence, A is the limiting reagent.
(iii) 1 atom of A combines with 1 molecule of B.
∴ All 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric and ther is no limiting reagent.
(iv) 1 mol of atom A combines with 1 mol of molecule B.
∴ 2.5 mol of B will combine with only 2.5 mol of A. and 2.5 mol of A will be left unreacted. Hence, B is the limiting reagent.
(v) 1 mol of atom A combines with 1 mol of molecule B.
∴ 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. Hence, A is the limiting reagent.
24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N_{2}(g) + H_{2}(g) → 2NH_{3}(g)
(i) Calculate the mass of ammonia produced if 2.00×10^{3}g dinitrogen reacts with 1.00×10^{3}g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution :
1 mole of dinitrogen (28g) reacts with 3 mole of dihydrogen (6g) to give 2 mole of ammonia (34g).
∴ 2000 g of N_{2 }will react with H_{2} = 6/28 × 200g = 428.6g. Thus, here N_{2 }is the limiting reagent while H_{2} is in excess.
28g of N_{2} produce 34g of NH_{3}.
∴2000g of N_{2} will produce = 34/28 × 2000g = 2428.57 g of NH_{3}.
(ii) N_{2} is the limiting reagent and H2 is the excess reagent. Hence, H_{2} will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1000g  428.6g = 571.4 g
25. How are 0.50 mol Na_{2}CO_{3} and 0.50 MNa_{2} CO_{3} different?
Solution :
Molar mass of Na_{2}CO_{3} = (2×23) +12.00+(3×16) = 106 g mol^{1}
∴0.50 mol Na_{2}CO_{3} means 0.50 ×106g = 53g
0.50 M Na2CO3 means 0.50 mol of Na2CO_{3} i.e. 53g of Na_{2}CO_{3} are present in 1litre of the solution.
26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Solution :
Dihydrogen gas reacts with dioxygen gas as,
2H_{2}(g) + O_{2}(g) → 2H_{2}O(g)
Thus, 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.
27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Solution :
(i) 1 pm = 10^{12} m
28.7 pm = 28.7 × 10^{12 }m = 2.87 × 10^{11 }m
(ii) 1 pm = 10^{12} m
∴15.15 pm = 15.15 × 10^{12 }m = 1.515 × 10^{11 }m
(iii) 1 mg = 10^{3} g
25365 mg = 2.5365 × 104×10^{3} g
Now,
1 g = 10^{3} kg
2.5365×10 g = 2.5365 × 10×10^{3} kg
∴25365 mg = 2.5365 × 10^{2} kg
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 Cl_{2} (g)
Solution :
(i) 1 g Au = 1/197 mol = 1/197 × 6.022×10^{23} atoms
(ii) 1 g Na = 1/23 mol = 1/23 × 6.022×10^{23 }atoms
(iii) 1 g Li = 1/7 mol = 1/7 × 6.022×10^{23} atoms
(iv) 1 g Cl2 = 1/71 mol = 1/71 × 6.022×10^{23} atoms
Thus, 1 g of Li has the largest number of atoms.
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).
Solution :
Mole fraction of C_{2}H_{5}OH = No. of moles of C_{2}H_{5}OH /No. of moles of solution
nC_{2}H_{5}OH = n(C_{2}H_{5}OH) / (C_{2}H_{5}OH) + n(H_{2}O) = 0.040 (Given) ... 1
We have to find the number of moles of ethanol in 1L of the solution but the solution is dilute. Therefor, water is approx. 1L.
No. of moles in 1L of water = 1000g /18g mol^{1} = 55.55 moles
Substituting n(H_{2}O) = 55.55 in equation 1
n(C2H5OH) / (C_{2}H_{5}OH) + 55.55 = 0.040
⇒ 0.96n (C_{2}H_{5}OH) = 55.55 × 0.040
⇒ n(C_{2}H_{5}OH) = 2.31 mol
Hence, molarity of the solution = 2.31M
30. What will be the mass of one 12C atom in g ?
Solution :
1 mol of 12C atoms = 6.022 ×10^{23} atoms = 12g
∴ Mass of 1 atom 12C = 12 /6.022 ×10^{23} g = 1.9927× 10^{23} g
31. How many significant figures should be present in the answer of the following calculations?
(i) 0.02856 × 298.15 × 0.112 /0.5785
(ii) 5 × 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Solution :
(i) Least precise term i.e. 0.112 is having 3 significant digits.
∴ There will be 3 significant figures in the calculation.
(ii) 5.364 is having 4 significant figures.
∴ There will be 4 significant figures in the calculation.
(iii) Least number of decimal places in each term is 4.
∴ There will be 4 significant figures in the calculation.
32. Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:
Isotope Isotopic molar mass Abundance
36Ar 35.96755 g mol^{1} 0.337%
38Ar 37.96272 g mol^{1 } 0.063%
40Ar 39.9624 g mol^{1} 99.600%
Solution :
Molar mass of Ar = ∑piAi
= (0.00337 × 35.96755 )+ (0.00063 × 37.96272 )+(0.99600 × 39.9624 ) = 39.948 g mol^{1}
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.
Solution :
(i) 1 mol of Ar = 6.022 × 10^{23}atoms
∴ 52 mol of Ar = 52 × 6.022×10^{23}atoms = 3.131 × 10^{25}atoms
(ii) 1 atom of He = 4 u of He
4 u of He = 1 Atom of He
∴ 52 u of He = 1/4 × 52 = 13 atoms
(iii) 1 mol of He = 4 g = 6.022 × 10^{23}atoms
∴ 52 g of He = (6.022 × 10^{23}/4) × 52 atoms = 7.8286 × 10^{24}atoms
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, and (iii) molecular formula.
Solution :
Amount of carbon in 3.38 g of CO_{2} = 12/44 × 3.38 g = 0.9218 g
Amount of hydrogen in 0.690 g H_{2}O = 2/18 × 0.690 g = 0.0767 g
The compound contains only C and H, therefore total mass of the compound = 0.9218 + 0.0767 = 0.9985 g
% of C in the compound = (0.9218 /0.9985 )×100 = 92.32
% of H in the compound = (0.0767 /0.9985 )×100 = 7.68
(i) Calculation of empirical formula,
Moles of carbon in the compound = 92.32/12 = 7.69
Moles of hydrogen in the compound = 7.68/1 = 7.68
Simplest molar ratio = 7.69 : 7.68 = 1(approx)
∴ Empirical formula CH
(ii) 10.0 L of the gas at STP weigh = 11.6 g
∴ 22.4 L of the gas at STP = 11.6/10.0 × 22.4 = 25.984 = 26 (approx)
∴ Molar mass of gass = 26 g mol1
(iii) Mass of empirical formula CH = 12+1 = 13
∴ n = Molecular Mass/Empirical Formula = 26/13 = 2
∴ Molecular Formula = C_{2}H_{2}
35. Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction, CaCO_{3} (s) + 2HCl (aq) → CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)
What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl?
Solution :
1000 mL of 0.75 M HCl have 0.75 mol of HCl = 0.75×36.5 g = 24.375 g
∴ Mass of HCl in 25mL of 0.75 M HCl = 24.375/ 1000 × 25 g = 0.6844 g
From the given chemical equation,
CaCO_{3} (s) + 2HCl (aq) → CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)
2 mol of HCl i.e. 73 g HCl react completely with 1 mol of CaCO_{3} i.e. 100g
∴ 0.6844 g HCl reacts completely with CaCO_{3} = 100/73 × 0.6844 g = 0.938 g
36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction
4HCl (aq) + MnO_{2}(s) → 2H_{2}O(l) + MnCl2(aq) + Cl_{2}(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution :
1 mol of MnO_{2} = 55+32 g = 87 g
87 g of MnO_{2} react with 4 moles of HCl i.e. 4×36.5 g = 146 g of HCl.
∴ 5.0 g of MnO_{2} will react with HCl = 146/87×5.0 g = 8.40 g.