Question

balancing and types of reactions practice
balance the equation:
□al₂(so₄)₃ + □ca(oh)₂ → □al(oh)₃ + □caso₄
type of reaction:

Explanation:

Step1: Balance Al atoms

Left side has 2 Al (in $\ce{Al_{2}(SO_{4})_{3}}$), right side has 1 Al (in $\ce{Al(OH)_{3}}$). So put 2 in front of $\ce{Al(OH)_{3}}$. Now the equation is: $\ce{Al_{2}(SO_{4})_{3} + \square Ca(OH)_{2} -> 2 Al(OH)_{3} + \square CaSO_{4}}$

Step2: Balance $\ce{SO_4^{2-}}$ ions

Left side has 3 $\ce{SO_4^{2-}}$ (in $\ce{Al_{2}(SO_{4})_{3}}$), so put 3 in front of $\ce{CaSO_{4}}$. Now: $\ce{Al_{2}(SO_{4})_{3} + \square Ca(OH)_{2} -> 2 Al(OH)_{3} + 3 CaSO_{4}}$

Step3: Balance Ca atoms

Right side has 3 Ca (in $\ce{CaSO_{4}}$), so put 3 in front of $\ce{Ca(OH)_{2}}$. Now: $\ce{Al_{2}(SO_{4})_{3} + 3 Ca(OH)_{2} -> 2 Al(OH)_{3} + 3 CaSO_{4}}$

Step4: Check OH and other atoms

Left side: 3 $\ce{Ca(OH)_{2}}$ has 6 OH, right side: 2 $\ce{Al(OH)_{3}}$ has 6 OH. Al: 2 on left, 2 on right. $\ce{SO_4^{2-}}$: 3 on left, 3 on right. Ca: 3 on left, 3 on right. Balanced.

Step5: Determine reaction type

This is a double - displacement reaction (ions exchange: $\ce{Al^{3+}}$ with $\ce{OH^-}$, $\ce{Ca^{2+}}$ with $\ce{SO_4^{2-}}$) and also a precipitation reaction (forms $\ce{Al(OH)_{3}}$ or $\ce{CaSO_{4}}$ as precipitate), but the main type here is double - replacement (double - displacement).

Answer:

Balanced equation: $\boldsymbol{\ce{1 Al_{2}(SO_{4})_{3} + 3 Ca(OH)_{2} -> 2 Al(OH)_{3} + 3 CaSO_{4}}}$
Type of reaction: Double - displacement (or Double - replacement)