(c) assertion is true but reason is false

CH_4(g) + 2O₂ \(\rightarrow\) CO₂ + 2H₂O
16g                          (2x18)g
As per stoichiometric equation,
16 g of methane produces 36 g of H₂O
\(\therefore\) 32 g of methane will produce = \(\frac { 36 }{ 16 } \times 32=72\) g of water.

At the anode, loss of electron takes place (ie.,) oxidation occurs. Hence called as oxidation electrode.
At the cathode, the gain of electrons takes place (ie.,) reduction occurs. Hence called as reduction electrode.

In a redox reaction, the total number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidising agent.

(i) Oxygen is being added to Cu, (ie.,) Cu₂S is oxidised to Cu₂O and the other reactant O₂ is getting reduced
(ii) Cu₂S is the reducing agent.
O₂ is an oxidising agent.

1) The unit of molecular mass is atomic mass unit [amu]. The unit of molar mass is gram per mole.
2) Molecular mass is the mass of one molecule while molar mass is the mass of one mole of molecules (6.022 x 10²³)
(i) Molecular mass of CO₂ = 1(C) + 2(0) = 12 + 32 = 44 amu
or 7.304 x 10^-23 g
(ii) Molar mass of CO₂ = 44 g mol^-1.

Molar mass = density x Molar volume
= 1.965 x 2.24 x 10^-2
= 4.4016 x 10^-2 kg/mol
= 4.4016 x 10^-2 x 10³ g/mol
= 44.016 g/mol

Step - 1 : To find atoms undergoing change in O.N.
\(\overset { +1+5-2 }{ KIO_{ 3 } } +\overset { +4-2 }{ SO_{ 2 } } +\overset { +1-2 }{ H_{ 2 }O } \rightarrow \overset { +1+1+6-2 }{ KHSO_{ 4 } } +\overset { +1+6-2 }{ H_{ 2 }SO_{ 4 } } +\overset { 0 }{ { I }_{ 2 } } \)
Step - 2 : To find the total decrease and increase in O.N.
KIO₃ ⟶  I₂ (decrease of 5 units per atom)
SO₂ ⟶ H₂SO₄ (increase of 2 units per atom)
Total decrease = 5 x 2 = 10
Total increase = 2 x 5 = 10
Step - 3 : To balance the total decrease and in O.N. increase, multiply KIO₃ by 2 and SO₂ by 5.
2 KIO₃ + 5SO₂ + H₂O ⟶ KHSO₄ + H₂SO₄ + I₂
Step - 4 : To balance all atoms other than 'O' and 'H'
2KIO₃ + 5SO₂ + H2O ⟶ 2KHSO₄ + 3H₂SO₄ + I₂
Step - 5 : To balance 'O' atoms
2 KIO₃ + 5SO₂ + 4H₂O ⟶ 2KHSO₄ + 3H₂SO₄ + I₂
Hydrogen atoms balance by themselves.
Hence, the balanced equations is
2KIO₃ + 5SO₂ + 4H₂O ⟶ 2KHSO₄ + 3H₂SO₄ + I₂

Step-1: To find atoms undergoing change in O.N
\(\overset { 0 }{ P } +\overset { +1\quad +5 }{ HNO_{ 3 } } \rightarrow \overset { +1+5-2 }{ HPO_{ 3 } } +\overset { +1-2 }{ NO } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find total decrease and increase in O.N.
P ⟶ HPO₃ (increase in O.N. of 5 units per atom)
HNO₃ ⟶ NO (decrease in O.N. of3 units per atom)
Total decrease 5 x 3 = 15
Total increase 3 x 5 = 15
Step-3: To balance the total increase and decrease in the equation, by multiplying P by 3 and HNO₃ by 5.
3P + 5HNO3  ⟶ HPO₃ + NO + H₂O
Step-4: To balance all atoms other than 'O' and 'H'
3P + 5HNO₃ ⟶ 3HPO₃ + 5NO + H₂O
Step-5: To balance by oxygen atoms
Oxygen and hydrogen atoms balance by themselves.
Hence the balanced equation is 3P + 5HNO3 ⟶ 3HPO₃ + 5 NO + H₂O

2Al + Fe₂O₃ \(\longrightarrow \)  Al₂O₃ + 2Fe

Molar mass of Al₂O₃ format = 6 mol x 102 g mol^-1 = 612 g
[  Al₂O₃ : (2 x 27) + 3(16) = 54 + 48 = 102] = 612 g
Excess reagent = Fe₂O₃
Amount of excess reagent left at the end of the reaction = 1 mol x 160 g mol^-1
= 160g [ Fe₂O₃ : (2 x 56) + (3 x 16) = 112 + 48 = 160] = 160 g

mass of one atom  = 6.645  x 10^-23 g
\(\therefore\) mass of 1 mole of an atom = 6.645  x 10^-23 g x 6.022 x 10²³ = 40 g
\(\therefore\) number of moles of element in 0.320 kg = \(\frac { 1\quad mole }{ 40g } \times 0.320kg\)
= \(\frac { 1mol\times 320g }{ 40g } \)
= 8 mol