Question

balancing and types of reactions practice
balance the equation:
$square$ca(oh)$_2$ + $square$h$_2$so$_4$ $
ightarrow$ $square$caso$_4$ + $square$h$_2$o
type of reaction:
$square$

Explanation:

Step1: Analyze Calcium (Ca) atoms

On the left side, we have 1 Ca atom in $\ce{Ca(OH)2}$. On the right side, we have 1 Ca atom in $\ce{CaSO4}$. So Ca is already balanced.

Step2: Analyze Sulfur (S) atoms

On the left side, we have 1 S atom in $\ce{H2SO4}$. On the right side, we have 1 S atom in $\ce{CaSO4}$. So S is already balanced.

Step3: Analyze Hydrogen (H) and Oxygen (O) atoms (from hydroxide and water)

On the left side, in $\ce{Ca(OH)2}$, we have 2 O atoms from $\ce{OH-}$ and 2 H atoms from $\ce{OH-}$. In $\ce{H2SO4}$, we have 4 O atoms and 2 H atoms. On the right side, in $\ce{CaSO4}$, we have 4 O atoms. In $\ce{H2O}$, we have 1 O atom and 2 H atoms per molecule.
Let's look at the $\ce{OH-}$ group. In $\ce{Ca(OH)2}$, there are 2 $\ce{OH-}$ groups, which means 2 H and 2 O. To balance the H and O from the hydroxide, we need 2 $\ce{H2O}$ molecules because each $\ce{H2O}$ has 2 H and 1 O. Wait, let's check the number of H atoms.
Left side H: 2 (from $\ce{Ca(OH)2}$) + 2 (from $\ce{H2SO4}$) = 4.
Right side H: 2 per $\ce{H2O}$ molecule. So we need 2 $\ce{H2O}$ molecules to have 4 H atoms.
Now check O atoms:
Left side O: 2 (from $\ce{Ca(OH)2}$) + 4 (from $\ce{H2SO4}$) = 6.
Right side O: 4 (from $\ce{CaSO4}$) + 2 (from 2 $\ce{H2O}$) = 6. Perfect.

So the coefficients are: 1 for $\ce{Ca(OH)2}$, 1 for $\ce{H2SO4}$, 1 for $\ce{CaSO4}$, and 2 for $\ce{H2O}$.

Step4: Determine the type of reaction

This is a reaction between an acid ($\ce{H2SO4}$) and a base ($\ce{Ca(OH)2}$) to form a salt ($\ce{CaSO4}$) and water ($\ce{H2O}$). So it's a neutralization reaction, which is a type of double - displacement reaction (specifically acid - base neutralization).

Answer:

The balanced equation is $\boldsymbol{1}\ce{Ca(OH)2 + 1H2SO4
ightarrow 1CaSO4 + 2H2O}$ and the type of reaction is Neutralization (Double - Displacement).