Question

write and balance the following chemical reactions. then identify each reaction type.

1. potassium metal reacts with liquid water yielding a solution of potassium hydroxide and hydrogen gas.
2. chlorine gas reacts with aqueous potassium bromide yielding a solution of potassium chloride and bromine gas.

Explanation:

Step1: Write the un - balanced equation for the first reaction

The un - balanced equation is \(K + H_2O
ightarrow KOH + H_2\).

Step2: Balance the first equation

Count the number of atoms. For hydrogen: on the left - hand side (LHS) we have \(2\) in \(H_2O\), on the right - hand side (RHS) we have \(1\) in \(KOH\) and \(2\) in \(H_2\). For potassium: \(1\) on LHS and \(1\) on RHS.
Multiply \(K\) by \(2\), \(H_2O\) by \(2\), \(KOH\) by \(2\). The balanced equation is \(2K+2H_2O = 2KOH + H_2\).
This is a single - displacement reaction because potassium (\(K\)) displaces hydrogen (\(H\)) from water (\(H_2O\)).

Step3: Write the un - balanced equation for the second reaction

The un - balanced equation is \(Cl_2+KBr
ightarrow KCl + Br_2\).

Step4: Balance the second equation

Count the number of atoms. For chlorine: \(2\) on LHS and \(1\) on RHS. For bromine: \(1\) on LHS and \(2\) on RHS.
Multiply \(KBr\) by \(2\) and \(KCl\) by \(2\). The balanced equation is \(Cl_2 + 2KBr=2KCl+Br_2\).
This is a single - displacement reaction because chlorine (\(Cl_2\)) displaces bromine (\(Br\)) from potassium bromide (\(KBr\)).

Answer:

1. Balanced equation: \(2K + 2H_2O=2KOH + H_2\), Reaction type: Single - displacement reaction.
2. Balanced equation: \(Cl_2+2KBr = 2KCl+Br_2\), Reaction type: Single - displacement reaction.