Balance each of the equations below, and identify the type of reaction in each case:

There are 4 Al on the left, so place a coefficient of 2 in front of $\text{Al}_2\text{O}_3$.

There are 3 C on the left, so place a coefficient of 3 in front of $\text{CO}$.

Left: $3 \times 3 = 9$. Right: $(2 \times 3) + (3 \times 1) = 6 + 3 = 9$. All atoms are balanced.
Balanced Equation: $\boxed{\text{Al}_4(\text{CO}_3)_3 \rightarrow 2\text{Al}_2\text{O}_3 + 3\text{CO}}$
Type of Reaction: Decomposition reaction (one compound breaks down into simpler substances).

Reactants: Ba: 1, O: $2+4=6$, H: 2, S: 1
Products: Ba: 1, O: $4+2=6$, H: 2, S: 1
All atoms are already balanced.
Balanced Equation: $\boxed{\text{BaO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + \text{H}_2\text{O}_2}$
Type of Reaction: Double displacement reaction (ions are exchanged between two compounds).

There are 6 O on the left, so place a coefficient of 3 in front of $\text{O}_2$.

Ba: 1 on both sides. Cl: 2 on both sides. All atoms are balanced.
Balanced Equation: $\boxed{\text{Ba}(\text{ClO}_3)_2 \rightarrow \text{BaCl}_2 + 3\text{O}_2}$
Type of Reaction: Decomposition reaction (one compound breaks down into simpler substances).

Balance the oxidation and reduction changes. Cu goes from 0 to +2 (oxidized, change of +2). N goes from +5 in $\text{HNO}_3$ to +2 in $\text{NO}$ (reduced, change of -3). To balance electrons, we need 3 Cu and 2 NO.

On the right, there are $3 \times 2 = 6$ N in $\text{Cu}(\text{NO}_3)_2$ and 2 N in $\text{NO}$, totaling 8 N. So, place a coefficient of 8 in front of $\text{HNO}_3$.

There are 8 H on the left. On the right, we need 8 H, so place a coefficient of 4 in front of $\text{H}_2\text{O}$.

Left: $8 \times 3 = 24$. Right: $(3 \times 2 \times 3) + (2 \times 1) + (4 \times 1) = 18 + 2 + 4 = 24$. All atoms are balanced.
Balanced Equation: $\boxed{3\text{Cu} + 8\text{HNO}_3 \rightarrow 3\text{Cu}(\text{NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}}$
Type of Reaction: Redox reaction (oxidation states of elements change).

There is 1 B on the left and 3 B on the right, so place a coefficient of 3 in front of $\text{H}_3\text{BO}_3$.

Reactants: H: $3 \times 3 = 9$, O: $3 \times 3 = 9$.
Products: H: 3 (from $\text{H}_3\text{B}_3\text{O}_6$) + 2 (from $\text{H}_2\text{O}$) = 5. O: 6 (from $\text{H}_3\text{B}_3\text{O}_6$) + 1 (from $\text{H}_2\text{O}$) = 7.
We need $9-3=6$ more H and $9-6=3$ more O on the product side. This corresponds to 3 molecules of $\text{H}_2\text{O}$. So, change the coefficient of $\text{H}_2\text{O}$ to 3.

Reactants: H: 9, B: 3, O: 9.
Products: H: $3 + (3 \times 2) = 9$, B: 3, O: $6 + (3 \times 1) = 9$. All atoms are balanced.
Balanced Equation: $\boxed{3\text{H}_3\text{BO}_3 \rightarrow \text{H}_3\text{B}_3\text{O}_6 + 3\text{H}_2\text{O}}$
Type of Reaction: Decomposition reaction (specifically, a dehydration reaction where water is removed).

Let the coefficients be $a\text{P}_4 + b\text{I}_2 + c\text{H}_2\text{O} \rightarrow d\text{PH}_4\text{I} + e\text{H}_3\text{PO}_4$.
P: $4a = d+e$
I: $2b = d$
H: $2c = 4d+3e$
O: $c = 4e$

$2(4e) = 4d+3e \implies 8e = 4d+3e \implies 5e = 4d$.

The smallest integers for $d$ and $e$ that satisfy $5e=4d$ are $d=5$ and $e=4$.

From I: $2b = 5 \implies b = 2.5$.
From P: $4a = 5+4 \implies 4a = 9 \implies a = 2.25$.

$a = 2.25 \times 4 = 9$
$b = 2.5 \times 4 = 10$
$d = 5 \times 4 = 20$
$e = 4 \times 4 = 16$

$c = 4 \times 16 = 64$.
Balanced Equation: $\boxed{9\text{P}_4 + 10\text{I}_2 + 64\text{H}_2\text{O} \rightarrow 20\text{PH}_4\text{I} + 16\text{H}_3\text{PO}_4}$
Type of Reaction: Redox reaction (specifically, a disproportionation reaction where phosphorus is both oxidized and reduced).