i) urea [CO(NH₂)₂]
Mol.mass = 1 (C) + 2(N) + 4(H) + 1(0)
= 1(12) + 2(14) + 4(1) + 1(16)
 = 12 + 28 + 4 + 16 = 60  
ii) Acetone [CH₃ COCH₃]
Mol.mass = 3(C) + 6(H) + 1(0)
= 3(12) + 6(1) + 1(16)
= 36 + 6 + 16 = 58
iii) Boric Acid [H₃ BO₃]
Mol.mass = 3(H) + 1(B) + 3(0)
= 3(1) + 1(11) + 3(16)
= 3 + 11 + 48 = 62
iv) Sulphuric  Acid [H₂ SO₄]
Mol.mass = 2(H) + 1(S) + 4(0)
= 2(1) + 1(32) + 4(16)
= 2 + 32 + 64 = 98

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

Half reactions are:
\(\overset { +7 }{ M } { nO }_{ 4 }^{ - }\longrightarrow { Mn }^{ 2+ }\)
and \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }\)
(1) \(\Rightarrow \) \({ MnO }_{ 4 }^{ - }+{ 8H }^{ - }+5e^{ - }\longrightarrow { Mn }^{ 2+ }+{ 4H }_{ 2 }O\)
(2) \(\Rightarrow \) \({ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+{ 2e }^{ - }\)

ii) 

iii) 

iv) 

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.

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