\(\underset { 2\ volumes }{ { 2H }_{ { 2 }_{ (g) } } } +\underset { 1\ volumes }{ { O }_{ { 2 }_{ (g) } } } \rightarrow \underset { 2\ volumes }{ { 2H }_{ 2 }{ O }_{ (g) } } \)
Thus 2 volumes of H₂ reacts with 1 volume of O₂ to produce 2 volumes of H₂O (g)
\(\therefore\) 10 volumes of H₂ would react with 5 volumes of O₂ to produce 10 volumes of H₂O(g).
Thus 10 volumes of H₂O will be produced.

It is defined as the imaginary charge left on the atom when all other atoms of the compound have been removed in their usual oxidation states that are assigned according to set of rules.

The balanced chemical equation is
\(\mathrm{MgCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_{2}\)
Molar mass of MgCO₃ is 84 g mol^–1.
84 g MgCO₃ contain 24 g of Magnesium.
∴ 100 g of MgCO₃ contain

= 28.57 g Mg.
i.e. percentage of magnesium
= 28.57 %.
84 g MgCO₃ contain 12 g of carbon

∴ 100 g MgCO₃ contain

= 14.29 g of carbon.
∴ Percentage of carbon
= 14.29 %.
84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

= 57.14 g of oxygen.
∴ Percentage of oxygen
= 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure MgCO₃ on heating gives 44 g of CO_2.
∴ 1000 g of 90 % pure MgCO₃ gives
\(\frac{\text { wt. of } \mathrm{MgCO}_{3}}{84 \mathrm{~g}} \frac{\text { % Purities }}{100 \%} \frac{\text { wt.of } \mathrm{CO}_{2}}{44 \mathrm{~g}}\)
\(x=44 \times \frac{100}{84} \times \frac{90}{100}\)
= 471.43 g CO₂
= 0.471 kg CO₂

Empirical formula = C₂H₃O₃
Empirical formula mass = 24 + 3 + 48 = 75
Molecular mass = 2 x Vapour density
= 2 x 75 = 150
n = \({Molecular \ mass \over Empirical \ formula \ mass }={150\over 75}=2\)
Molecular formula = (Empirical formula)_n
\(\therefore\) Molecular formula = C₂H₃O₃ x 2
Molecular formula = C₄H₆O₆

half reaction \(\Rightarrow \) \({ S }_{ 2 }{ O }_{ 3 }^{ 2- }\longrightarrow { S }_{ 4 }{ O }_{ 6 }^{ 2- }\)
\({ I }_{ 2 }\longrightarrow { I }^{ - }\)
 

Empirical formula = Na₂ SH₂₀ O₁₄ 
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄ 10H₂O.

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.

Combination reaction

Al₂(SO₄)₃ + 3 BaCl₂ \(\longrightarrow\) 3 BaSO₄ + 2 AlCl₃
When 1 mole of aluminium sulphate reacts with barium chloride, 3 moles of BaSO₄ is precipitated.

The oxidation state of CI in ClO₂^- is +3. So, chlorine can get oxidised as well as reduced and can act as reductant and oxidant.
The disproportionation reaction of ClO₂^- is
\(\begin{matrix} +1 \\ 3Cl{ O }_{ 2 }^{ - } \end{matrix}\longrightarrow { Cl }^{ - }+\begin{matrix} +5 \\ Cl{ O }_{ 3 }^{ - } \end{matrix}\)
In CIO₄^- , CI is in its highest oxidation state, So it can only be an oxidant.

\({ H }_{ 2 }\overset { -1 }{ \underset { \overset { \uparrow }{ (ie\times 2) } }{ { O }_{ 2 } } } +{ \underset { \overset { \downarrow }{ Ie^{ - } } }{ Fe } }^{ 2 }+\rightarrow { Fe }^{ 3+ }+\overset { -2 }{ {H}_{2} O} \)
\({ H }_{ 2 }{ O }_{ 2 }+{ 2Fe }^{ 2+ }+\rightarrow Fe^{ 3+ }+{ H }_{ 2 }O\)
\(\Rightarrow \) \({ H }_{ 2 }{ O }_{ 2 }+{ 2Fe }^{ 2+ }+{ 2 }H^{ + }\rightarrow 2Fe^{ 3+ }+{ 2H }_{ 2 }O\)