The balanced equation for the formation of HCI is,
H₂(g) + CI₂(g) \(\rightarrow\) 2 HCI (g)
As per the stoichiometric equation, under given conditions,
To produce 2 moles of HCI, 1 mole of chlorine gas is required.
To produce 44.8 litres of HCI, 22.4 litres of chlorine gas are required.
\(\therefore\) To produce 11.2 litres of HCI,

= 5.6 litres of chlorine are required.

The balanced chemical equation is,
\({ CaCO }_{ 3 }\left( s \right) \overset { \triangle }{ \longrightarrow } { CaO }_{ (s) }+{ CO }_{ 2 }(g)\)
As per the stoichiometric equation,
1 mole (100g) CaC0₃ on heating produces 1 mole CO₂

At STP, 1 mole of CO₂ occupies a volume of 22.7 liters
At STP, 50 g of CaCO₃ on heating produces,

= 11.35 liters of CO₂

The balanced stoichiometric equation for the formation of ammonia is
N₂(g) + 3H₂ (g) \(\rightarrow\) 2NH₃ (g)
As per the stoichiometric equation,
To produce 2 moles of ammonia, 3 moles of hydrogen are required.
\(\therefore\)  to produce 10 moles of ammonia,

= 15 moles of hydrogen are required.

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.

The balanced equation for the combustion of ethane
C₂H₆ + \(\frac { 7 }{ 2 } \)O₂ \(\longrightarrow \) 2CO₂ +3H₂O
2C₂H₆ + 7O₂ \(\longrightarrow \) 4CO₂ + 6H₂O
To produce 4 moles of CO₂, 2 moles of ethane is required
To produce 1 moles (44 g) of CO₂ required
Number of moles of ethane
\(=\frac{2 \mathrm{~mol} \text { ethane }}{4 \not \mathrm{molCO}_{2}} \times 1 \not \mathrm{molCO}_{2}\)
= \(\frac { 1 }{ 2 } mole\quad of\quad ethane\)
= 0.5 mole of ethane

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.

Question	Number of moles of reactants allowed to react			Number of moles reactants consumed during reaction			Limiting reagent
	x	y	z2	x	y	z2
(a)	200 atoms	200 atoms	50 molecules	50 atoms	50 atoms	50 molecules	z2

ii) 1 mol of x + 1 mol of y + 3 mol of z₂

Question	Number of moles of reactants allowed to react			Number of moles reactants consumed during reaction			Limiting reagent
	x	y	z2	x	y	z2
(a)	1 mol	1 mol	3 mol	1 mol	1 mol	1 mol	x and y

ii) 50 atoms of x + 25 atoms of y + 50 molecules of z₂

 

Question	Number of moles of reactants allowed to react			Number of moles reactants consumed during reaction			Limiting reagent
	x	y	z2	x	y	z2
(a)	50 atom	25 atom	50 molecules	25 atom	25 atom	25 molecules	y

iv)  2.5 mol of x + 5 mol of y + 5 mol of z₂

 

Question	Number of moles of reactants allowed to react			Number of moles reactants consumed during reaction			Limiting reagent
	x	y	z2	x	y	z2
(a)	2.5 mol	5 mol	5 mol	2.5 mol	2.5 mol	2.5 mol	x

Element	Percentage	Atomic mass	Relative number of atoms	simple ratio	Whole no
C	76.6	12	\(\frac { 76.6 }{ 12 } =6.38\)	\(\frac { 6.38 }{ 1.06 } =6\)	6
H	6.38	1	\(\frac { 6.38 }{ 1 } =6.38\)	\(\frac { 6.38 }{ 1.06 } =6\)	6
0	17.02	16	\(\frac { 17.02 }{ 16 } =1.06\)	\(1.06\frac { 1.06 }{ 1.06 } =1\)	1

Empirical Formula = C₆ H₆O
n = \(\frac { molar\ mass }{ calculated\ eprirical\ formula\ mass } \)
= \(\frac { 2\times \ vapour\ density }{ 94 } \frac { 2\times 47 }{ 94 } =1,\) 
Molecular formula (C₆H₆O)  x 1 = C₆H₆O.

Compound	Molecular formula	Empirical Formula
Fructose	C6 H12O6	C H2O
Caffeine	C8 H10 N4 O2	C4 H5 N2O

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