Answer
Balancing the equations by Inspection method
(a) $4Al + 6H_{2}SO4 →2Al_{2}(SO{4})_{3} + 6H_{2}$
(b) $2C_{5}H_{10} + 15O_{2} → 10CO_{2} + 10H_{2}O$
(c) $6Li + N_{2} → 2Li_{3}N$
(d) $Ba(ClO_{4}){2} →BaCl_{2} + 4O_{2}$
(e) $2C_{2}H_{6}O + 6O_{2} → 4CO_{2} + 6H_{2}O$
Work Step by Step
In the equations, we need to balance every atoms on left and right side of the equation. So, each atom should be counted and balanced by multiplying with a suitable number to get equal number of that atom on each side.
(a) $Al + H_{2}SO_{4} →Al_{2}(SO_{4})_{3} + H_{2}$
Here, atoms on LHS and RHS are as follows;
LHS:
1Al; 2H; 4O and 1S
RHS:
2Al; 2H; 12O and 3S
Al is balanced by adding 4 on LHS and 2 on RHS.
O and S are balanced by adding 6 on LHS to $H_{2}SO_{4}$
H is balanced by adding 6 on RHS to $H_{2}$.
The final balanced equation is;
$4Al + 6H_{2}SO_{4} →2Al_{2}(SO_{4})_{3} + 6H_{2}$
(b) $C_{5}H_{10} + O_{2} → CO_{2} + H_{2}O$
On LHS;
5C, 10H and 2O
on RHS;
1C, 3O and 2H atoms
by multiplying $C_{5}H_{10}$ with 2 and $O_{2} $ with 15 balances C, H and O on both sides.
The final balanced equation is;
$2C_{5}H_{10} + 15O_{2} → 10CO_{2} + 10H_{2}O$
(c) $Li + N_{2} → Li_{3}N$
on LHS there are; 1Li, 2N atoms
on RHS there are 3Li and 1N atoms.
By multiplying 6 to Li on LHS and 2 to $Li_{3}N$ balances the atoms.
The final balanced equation is;
$6Li + N_{2} → 2Li_{3}N$
(d) $Ba(ClO_{4})_{2} →BaCl_{2} + O_{2}$
Here Ba, and Cl atoms are already balanced on both sides. Oxygen atom can be balanced by adding 4 to $O_{2}$ on RHS.
The final balanced equation is;
$Ba(ClO_{4})_{2} →BaCl_{2} + 4O_{2}$
(e) $C_{2}H_{6}O + O_{2} → CO_{2} + H_{2}O$
On LHS; 2C, 6H and 3 O atoms;
On RHS; 1C, 3 O and 2 H atoms.
By adding 2 to $C_{2}H_{6}O$, 4 to $CO_{2}$ and 6 to $H_{2}O$ balances C and H atoms. Oxygen can be balanced by adding 6 to $O_{2}$ on LHS
The final balanced equation is;
$2C_{2}H_{6}O + 6O_{2} → 4CO_{2} + 6H_{2}O$
