Answer :
Certainly! Let's go through each of the chemical reactions and balance them step by step.
### Part a. Potassium hydroxide reacts with phosphoric acid
The unbalanced equation is:
[tex]\[ \text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow \text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
Steps to balance:
1. Count all atoms on both sides:
- Left: [tex]\( \text{K}=1 \)[/tex], [tex]\( \text{O}=5 \)[/tex] (1 in KOH and 4 in H₃PO₄), [tex]\( \text{H}=6 \)[/tex] (1 in KOH and 5 in H₃PO₄), [tex]\( \text{P}=1 \)[/tex]
- Right: [tex]\( \text{K}=3 \)[/tex], [tex]\( \text{O}=5 \)[/tex] (1 in H₂O and 4 in K₃PO₄), [tex]\( \text{H}=2 \)[/tex] (in H₂O), [tex]\( \text{P}=1 \)[/tex]
2. To balance potassium (K), put a coefficient of 3 before KOH:
[tex]\[ 3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 3\text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
3. Verify all other atoms are balanced. Now, both sides have 3 K, 6 H, 4 O in the polyatomic ions and 3 O in H₂O, and 1 P.
Final balanced equation:
[tex]\[ 3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 3\text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
### Part b. Silver nitrate reacts with sodium chloride
The unbalanced equation is:
[tex]\[ \text{AgNO}_3 (aq) + \text{NaCl} (s) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \][/tex]
Steps to balance:
1. Count the atoms on both sides:
- Left: [tex]\( \text{Ag}=1 \)[/tex], [tex]\( \text{N}=1 \)[/tex], [tex]\( \text{O}=3 \)[/tex], [tex]\( \text{Na}=1 \)[/tex], [tex]\( \text{Cl}=1 \)[/tex]
- Right: [tex]\( \text{Ag}=1 \)[/tex], [tex]\( \text{N}=1 \)[/tex], [tex]\( \text{O}=3 \)[/tex], [tex]\( \text{Na}=1 \)[/tex], [tex]\( \text{Cl}=1 \)[/tex]
2. All elements are already balanced.
Final balanced equation:
[tex]\[ \text{AgNO}_3 (aq) + \text{NaCl} (s) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \][/tex]
### Part c. Calcium hydroxide reacts with phosphoric acid
The unbalanced equation is:
[tex]\[ \text{Ca(OH)}_2 (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow \text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
Steps to balance:
1. Count all atoms on both sides:
- Left: [tex]\( \text{Ca}=1 \)[/tex], [tex]\( \text{O}=6 \)[/tex] (2 in Ca(OH)₂ and 4 in H₃PO₄), [tex]\( \text{H}=5 \)[/tex] (2 in Ca(OH)₂ and 3 in H₃PO₄), [tex]\( \text{P}=1 \)[/tex]
- Right: [tex]\( \text{Ca}=3 \)[/tex], [tex]\( \text{O}=8 \)[/tex] (in Ca₃(PO₄)₂), [tex]\( \text{H}=2 \)[/tex] (in H₂O), [tex]\( \text{P}=2 \)[/tex]
2. To balance calcium (Ca), put a coefficient of 3 before Ca(OH)₂:
[tex]\[ 3\text{Ca(OH)}_2 (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
3. To balance phosphorous (P), put a coefficient of 2 before H₃PO₄:
[tex]\[ 3\text{Ca(OH)}_2 (aq) + 2\text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
4. Verify other atoms:
- Left: [tex]\( \text{Ca}=3 \)[/tex], [tex]\( \text{O}=12 \)[/tex] (6 in 3Ca(OH)₂, 8 in 2H₃PO₄), [tex]\( \text{H}= 12 \)[/tex] (6 in 3Ca(OH)₂, 12 in 2H₃PO₄), [tex]\( \text{P}=2 \)[/tex]
Final balanced equation:
[tex]\[ 3\text{Ca(OH)}_2 (aq) + 2\text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
### Part d. Iron(II) sulfide reacts with hydrochloric acid
The unbalanced equation is:
[tex]\[ \text{FeS} (s) + \text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
Steps to balance:
1. Count all atoms on both sides:
- Left: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=1 \)[/tex], [tex]\( \text{Cl}=1 \)[/tex]
- Right: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=2 \)[/tex], [tex]\( \text{Cl}=2 \)[/tex]
2. To balance both H and Cl, put a coefficient of 2 before HCl:
[tex]\[ \text{FeS} (s) + 2\text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
3. Verify all other atoms:
- Left: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=2 \)[/tex], [tex]\( \text{Cl}=2 \)[/tex]
- Right: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=2 \)[/tex], [tex]\( \text{Cl}=2 \)[/tex]
Final balanced equation:
[tex]\[ \text{FeS} (s) + 2\text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
In summary, the balanced equations are:
a. [tex]\[ 3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 3\text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
b. [tex]\[ \text{AgNO}_3 (aq) + \text{NaCl} (s) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \][/tex]
c. [tex]\[ 3\text{Ca(OH)}_2 (aq) + 2\text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
d. [tex]\[ \text{FeS} (s) + 2\text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
### Part a. Potassium hydroxide reacts with phosphoric acid
The unbalanced equation is:
[tex]\[ \text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow \text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
Steps to balance:
1. Count all atoms on both sides:
- Left: [tex]\( \text{K}=1 \)[/tex], [tex]\( \text{O}=5 \)[/tex] (1 in KOH and 4 in H₃PO₄), [tex]\( \text{H}=6 \)[/tex] (1 in KOH and 5 in H₃PO₄), [tex]\( \text{P}=1 \)[/tex]
- Right: [tex]\( \text{K}=3 \)[/tex], [tex]\( \text{O}=5 \)[/tex] (1 in H₂O and 4 in K₃PO₄), [tex]\( \text{H}=2 \)[/tex] (in H₂O), [tex]\( \text{P}=1 \)[/tex]
2. To balance potassium (K), put a coefficient of 3 before KOH:
[tex]\[ 3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 3\text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
3. Verify all other atoms are balanced. Now, both sides have 3 K, 6 H, 4 O in the polyatomic ions and 3 O in H₂O, and 1 P.
Final balanced equation:
[tex]\[ 3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 3\text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
### Part b. Silver nitrate reacts with sodium chloride
The unbalanced equation is:
[tex]\[ \text{AgNO}_3 (aq) + \text{NaCl} (s) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \][/tex]
Steps to balance:
1. Count the atoms on both sides:
- Left: [tex]\( \text{Ag}=1 \)[/tex], [tex]\( \text{N}=1 \)[/tex], [tex]\( \text{O}=3 \)[/tex], [tex]\( \text{Na}=1 \)[/tex], [tex]\( \text{Cl}=1 \)[/tex]
- Right: [tex]\( \text{Ag}=1 \)[/tex], [tex]\( \text{N}=1 \)[/tex], [tex]\( \text{O}=3 \)[/tex], [tex]\( \text{Na}=1 \)[/tex], [tex]\( \text{Cl}=1 \)[/tex]
2. All elements are already balanced.
Final balanced equation:
[tex]\[ \text{AgNO}_3 (aq) + \text{NaCl} (s) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \][/tex]
### Part c. Calcium hydroxide reacts with phosphoric acid
The unbalanced equation is:
[tex]\[ \text{Ca(OH)}_2 (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow \text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
Steps to balance:
1. Count all atoms on both sides:
- Left: [tex]\( \text{Ca}=1 \)[/tex], [tex]\( \text{O}=6 \)[/tex] (2 in Ca(OH)₂ and 4 in H₃PO₄), [tex]\( \text{H}=5 \)[/tex] (2 in Ca(OH)₂ and 3 in H₃PO₄), [tex]\( \text{P}=1 \)[/tex]
- Right: [tex]\( \text{Ca}=3 \)[/tex], [tex]\( \text{O}=8 \)[/tex] (in Ca₃(PO₄)₂), [tex]\( \text{H}=2 \)[/tex] (in H₂O), [tex]\( \text{P}=2 \)[/tex]
2. To balance calcium (Ca), put a coefficient of 3 before Ca(OH)₂:
[tex]\[ 3\text{Ca(OH)}_2 (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
3. To balance phosphorous (P), put a coefficient of 2 before H₃PO₄:
[tex]\[ 3\text{Ca(OH)}_2 (aq) + 2\text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
4. Verify other atoms:
- Left: [tex]\( \text{Ca}=3 \)[/tex], [tex]\( \text{O}=12 \)[/tex] (6 in 3Ca(OH)₂, 8 in 2H₃PO₄), [tex]\( \text{H}= 12 \)[/tex] (6 in 3Ca(OH)₂, 12 in 2H₃PO₄), [tex]\( \text{P}=2 \)[/tex]
Final balanced equation:
[tex]\[ 3\text{Ca(OH)}_2 (aq) + 2\text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
### Part d. Iron(II) sulfide reacts with hydrochloric acid
The unbalanced equation is:
[tex]\[ \text{FeS} (s) + \text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
Steps to balance:
1. Count all atoms on both sides:
- Left: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=1 \)[/tex], [tex]\( \text{Cl}=1 \)[/tex]
- Right: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=2 \)[/tex], [tex]\( \text{Cl}=2 \)[/tex]
2. To balance both H and Cl, put a coefficient of 2 before HCl:
[tex]\[ \text{FeS} (s) + 2\text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
3. Verify all other atoms:
- Left: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=2 \)[/tex], [tex]\( \text{Cl}=2 \)[/tex]
- Right: [tex]\( \text{Fe}=1 \)[/tex], [tex]\( \text{S}=1 \)[/tex], [tex]\( \text{H}=2 \)[/tex], [tex]\( \text{Cl}=2 \)[/tex]
Final balanced equation:
[tex]\[ \text{FeS} (s) + 2\text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
In summary, the balanced equations are:
a. [tex]\[ 3\text{KOH} (aq) + \text{H}_3\text{PO}_4 (aq) \rightarrow 3\text{H}_2\text{O} (l) + \text{K}_3\text{PO}_4 (aq) \][/tex]
b. [tex]\[ \text{AgNO}_3 (aq) + \text{NaCl} (s) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \][/tex]
c. [tex]\[ 3\text{Ca(OH)}_2 (aq) + 2\text{H}_3\text{PO}_4 (aq) \rightarrow 6\text{H}_2\text{O} (l) + \text{Ca}_3 (\text{PO}_4)_2 (aq) \][/tex]
d. [tex]\[ \text{FeS} (s) + 2\text{HCl} (aq) \rightarrow \text{H}_2\text{S} (g) + \text{FeCl}_2 (aq) \][/tex]
