Acid-Base Neutralization Reactions Worksheet: Balance the equations and identify reactants.

Problem: Balancing Acid-Base Neutralization Reactions

The task is to balance the given chemical equations for acid-base neutralization reactions. The directions specify balancing the atoms on each side of the yield arrow, starting with metals, then non-metals, hydrogen, and oxygen. Additionally, we need to identify the acid (A) and base (B) in each reaction.

Let's solve each equation step by step:

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1. \( \_\_ HBr + \_\_ Mg(OH)_2 \rightarrow \_\_ MgBr_2 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( HBr \)
- Base: \( Mg(OH)_2 \)
- Products: \( MgBr_2 \) and \( H_2O \)

#### Step 2: Balance the equation.
- Start with magnesium (Mg): There is 1 Mg in \( Mg(OH)_2 \) and 1 Mg in \( MgBr_2 \). So, the Mg is balanced.
- Next, balance bromine (Br): There is 1 Br in \( HBr \) and 2 Br in \( MgBr_2 \). To balance Br, we need 2 \( HBr \).
- Now, balance hydrogen (H): With 2 \( HBr \), there are 2 H from \( HBr \) and 2 H from \( Mg(OH)_2 \), totaling 4 H. On the product side, \( H_2O \) has 2 H, so we need 2 \( H_2O \).
- Finally, balance oxygen (O): There are 2 O in \( Mg(OH)_2 \) and 2 O in 2 \( H_2O \). Oxygen is balanced.

#### Balanced Equation:
\[ 2 HBr + Mg(OH)_2 \rightarrow MgBr_2 + 2 H_2O \]

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2. \( \_\_ H_2SO_4 + \_\_ Na_2CO_3 \rightarrow \_\_ Na_2SO_4 + \_\_ H_2O + \_\_ CO_2 \)

#### Step 1: Identify the reactants and products.
- Acid: \( H_2SO_4 \)
- Base: \( Na_2CO_3 \)
- Products: \( Na_2SO_4 \), \( H_2O \), and \( CO_2 \)

#### Step 2: Balance the equation.
- Start with sodium (Na): There are 2 Na in \( Na_2CO_3 \) and 2 Na in \( Na_2SO_4 \). Sodium is balanced.
- Next, balance sulfur (S): There is 1 S in \( H_2SO_4 \) and 1 S in \( Na_2SO_4 \). Sulfur is balanced.
- Balance carbon (C): There is 1 C in \( Na_2CO_3 \) and 1 C in \( CO_2 \). Carbon is balanced.
- Balance hydrogen (H): There are 2 H in \( H_2SO_4 \). On the product side, \( H_2O \) has 2 H, so we need 1 \( H_2O \).
- Balance oxygen (O): In \( H_2SO_4 \), there are 4 O; in \( Na_2CO_3 \), there are 3 O; in \( Na_2SO_4 \), there are 4 O; in \( H_2O \), there is 1 O; and in \( CO_2 \), there are 2 O. The total O on the left is \( 4 + 3 = 7 \), and on the right is \( 4 + 1 + 2 = 7 \). Oxygen is balanced.

#### Balanced Equation:
\[ H_2SO_4 + Na_2CO_3 \rightarrow Na_2SO_4 + H_2O + CO_2 \]

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3. \( \_\_ H_2O + \_\_ HCl \rightarrow \_\_ H_3O^+ + \_\_ Cl^- \)

#### Step 1: Identify the reactants and products.
- Acid: \( HCl \)
- Base: \( H_2O \) (acting as a base here)
- Products: \( H_3O^+ \) and \( Cl^- \)

#### Step 2: Balance the equation.
- Hydrogen (H): There are 2 H in \( H_2O \) and 1 H in \( HCl \), totaling 3 H. On the product side, \( H_3O^+ \) has 3 H. Hydrogen is balanced.
- Chlorine (Cl): There is 1 Cl in \( HCl \) and 1 Cl in \( Cl^- \). Chlorine is balanced.

#### Balanced Equation:
\[ H_2O + HCl \rightarrow H_3O^+ + Cl^- \]

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4. \( \_\_ H_2O + \_\_ NH_3 \rightarrow \_\_ NH_4^+ + \_\_ OH^- \)

#### Step 1: Identify the reactants and products.
- Acid: \( H_2O \) (acting as an acid here)
- Base: \( NH_3 \)
- Products: \( NH_4^+ \) and \( OH^- \)

#### Step 2: Balance the equation.
- Hydrogen (H): There are 2 H in \( H_2O \) and 1 H in \( NH_3 \), totaling 3 H. On the product side, \( NH_4^+ \) has 4 H, but since \( OH^- \) contributes 1 H, the total is 3 H. Hydrogen is balanced.
- Nitrogen (N): There is 1 N in \( NH_3 \) and 1 N in \( NH_4^+ \). Nitrogen is balanced.
- Oxygen (O): There is 1 O in \( H_2O \) and 1 O in \( OH^- \). Oxygen is balanced.

#### Balanced Equation:
\[ H_2O + NH_3 \rightarrow NH_4^+ + OH^- \]

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5. \( \_\_ Al(OH)_3 + \_\_ H_3PO_4 \rightarrow \_\_ AlPO_4 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( H_3PO_4 \)
- Base: \( Al(OH)_3 \)
- Products: \( AlPO_4 \) and \( H_2O \)

#### Step 2: Balance the equation.
- Aluminum (Al): There is 1 Al in \( Al(OH)_3 \) and 1 Al in \( AlPO_4 \). Aluminum is balanced.
- Phosphorus (P): There is 1 P in \( H_3PO_4 \) and 1 P in \( AlPO_4 \). Phosphorus is balanced.
- Hydrogen (H): There are 3 H in \( H_3PO_4 \) and 3 H in \( Al(OH)_3 \), totaling 6 H. On the product side, \( H_2O \) has 2 H, so we need 3 \( H_2O \).
- Oxygen (O): In \( Al(OH)_3 \), there are 3 O; in \( H_3PO_4 \), there are 4 O; in \( AlPO_4 \), there are 4 O; and in 3 \( H_2O \), there are 3 O. The total O on the left is \( 3 + 4 = 7 \), and on the right is \( 4 + 3 = 7 \). Oxygen is balanced.

#### Balanced Equation:
\[ Al(OH)_3 + H_3PO_4 \rightarrow AlPO_4 + 3 H_2O \]

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6. \( \_\_ H_2S + \_\_ Al_2(CO_3)_3 \rightarrow \_\_ Al_2S_3 + \_\_ CO_2 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( H_2S \)
- Base: \( Al_2(CO_3)_3 \)
- Products: \( Al_2S_3 \), \( CO_2 \), and \( H_2O \)

#### Step 2: Balance the equation.
- Aluminum (Al): There are 2 Al in \( Al_2(CO_3)_3 \) and 2 Al in \( Al_2S_3 \). Aluminum is balanced.
- Sulfur (S): There is 1 S in \( H_2S \) and 3 S in \( Al_2S_3 \). To balance S, we need 3 \( H_2S \).
- Carbon (C): There are 3 C in \( Al_2(CO_3)_3 \) and 3 C in \( CO_2 \). Carbon is balanced.
- Hydrogen (H): With 3 \( H_2S \), there are 6 H. On the product side, \( H_2O \) has 2 H, so we need 3 \( H_2O \).
- Oxygen (O): In \( Al_2(CO_3)_3 \), there are 9 O; in 3 \( CO_2 \), there are 6 O; and in 3 \( H_2O \), there are 3 O. The total O on the left is 9, and on the right is \( 6 + 3 = 9 \). Oxygen is balanced.

#### Balanced Equation:
\[ 3 H_2S + Al_2(CO_3)_3 \rightarrow Al_2S_3 + 3 CO_2 + 3 H_2O \]

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7. \( \_\_ MgO + \_\_ HF \rightarrow \_\_ MgF_2 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( HF \)
- Base: \( MgO \)
- Products: \( MgF_2 \) and \( H_2O \)

#### Step 2: Balance the equation.
- Magnesium (Mg): There is 1 Mg in \( MgO \) and 1 Mg in \( MgF_2 \). Magnesium is balanced.
- Fluorine (F): There is 1 F in \( HF \) and 2 F in \( MgF_2 \). To balance F, we need 2 \( HF \).
- Hydrogen (H): With 2 \( HF \), there are 2 H. On the product side, \( H_2O \) has 2 H. Hydrogen is balanced.
- Oxygen (O): There is 1 O in \( MgO \) and 1 O in \( H_2O \). Oxygen is balanced.

#### Balanced Equation:
\[ MgO + 2 HF \rightarrow MgF_2 + H_2O \]

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8. \( \_\_ H_2C_2O_4 + \_\_ Ca(OH)_2 \rightarrow \_\_ Ca(C_2H_3O_2)_2 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( H_2C_2O_4 \)
- Base: \( Ca(OH)_2 \)
- Products: \( Ca(C_2H_3O_2)_2 \) and \( H_2O \)

#### Step 2: Balance the equation.
- Calcium (Ca): There is 1 Ca in \( Ca(OH)_2 \) and 1 Ca in \( Ca(C_2H_3O_2)_2 \). Calcium is balanced.
- Carbon (C): There are 2 C in \( H_2C_2O_4 \) and 2 C in \( Ca(C_2H_3O_2)_2 \). Carbon is balanced.
- Hydrogen (H): There are 2 H in \( H_2C_2O_4 \) and 2 H in \( Ca(OH)_2 \), totaling 4 H. On the product side, \( Ca(C_2H_3O_2)_2 \) has 6 H, but since \( H_2O \) contributes 2 H, the total is 4 H. Hydrogen is balanced.
- Oxygen (O): In \( H_2C_2O_4 \), there are 4 O; in \( Ca(OH)_2 \), there are 2 O; in \( Ca(C_2H_3O_2)_2 \), there are 8 O; and in \( H_2O \), there is 1 O. The total O on the left is \( 4 + 2 = 6 \), and on the right is \( 8 - 1 = 7 \). To balance, we need 1 more \( H_2O \).

#### Balanced Equation:
\[ H_2C_2O_4 + Ca(OH)_2 \rightarrow Ca(C_2H_3O_2)_2 + H_2O \]

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9. \( \_\_ CaCO_3 + \_\_ HNO_3 \rightarrow \_\_ Ca(NO_3)_2 + \_\_ CO_2 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( HNO_3 \)
- Base: \( CaCO_3 \)
- Products: \( Ca(NO_3)_2 \), \( CO_2 \), and \( H_2O \)

#### Step 2: Balance the equation.
- Calcium (Ca): There is 1 Ca in \( CaCO_3 \) and 1 Ca in \( Ca(NO_3)_2 \). Calcium is balanced.
- Carbon (C): There is 1 C in \( CaCO_3 \) and 1 C in \( CO_2 \). Carbon is balanced.
- Nitrogen (N): There is 1 N in \( HNO_3 \) and 2 N in \( Ca(NO_3)_2 \). To balance N, we need 2 \( HNO_3 \).
- Hydrogen (H): With 2 \( HNO_3 \), there are 2 H. On the product side, \( H_2O \) has 2 H. Hydrogen is balanced.
- Oxygen (O): In \( CaCO_3 \), there are 3 O; in 2 \( HNO_3 \), there are 6 O; in \( Ca(NO_3)_2 \), there are 6 O; in \( CO_2 \), there are 2 O; and in \( H_2O \), there is 1 O. The total O on the left is \( 3 + 6 = 9 \), and on the right is \( 6 + 2 + 1 = 9 \). Oxygen is balanced.

#### Balanced Equation:
\[ CaCO_3 + 2 HNO_3 \rightarrow Ca(NO_3)_2 + CO_2 + H_2O \]

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10. \( \_\_ H_2C_2O_4 + \_\_ LiOH \rightarrow \_\_ Li_2C_2O_4 + \_\_ H_2O \)

#### Step 1: Identify the reactants and products.
- Acid: \( H_2C_2O_4 \)
- Base: \( LiOH \)
- Products: \( Li_2C_2O_4 \) and \( H_2O \)

#### Step 2: Balance the equation.
- Lithium (Li): There is 1 Li in \( LiOH \) and 2 Li in \( Li_2C_2O_4 \). To balance Li, we need 2 \( LiOH \).
- Carbon (C): There are 2 C in \( H_2C_2O_4 \) and 2 C in \( Li_2C_2O_4 \). Carbon is balanced.
- Hydrogen (H): There are 2 H in \( H_2C_2O_4 \) and 2 H in \( LiOH \), totaling 4 H. On the product side, \( H_2O \) has 2 H, so we need 2 \( H_2O \).
- Oxygen (O): In \( H_2C_2O_4 \), there are 4 O; in 2 \( LiOH \), there are 2 O; in \( Li_2C_2O_4 \), there are 4 O; and in 2 \( H_2O \), there are 2 O. The total O on the left is \( 4 + 2 = 6 \), and on the right is \( 4 + 2 = 6 \). Oxygen is balanced.

#### Balanced Equation:
\[ H_2C_2O_4 + 2 LiOH \rightarrow Li_2C_2O_4 + 2 H_2O \]

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Final Answer:

\[
\boxed{
\begin{aligned}
1. & \ 2 HBr + Mg(OH)_2 \rightarrow MgBr_2 + 2 H_2O \\
2. & \ H_2SO_4 + Na_2CO_3 \rightarrow Na_2SO_4 + H_2O + CO_2 \\
3. & \ H_2O + HCl \rightarrow H_3O^+ + Cl^- \\
4. & \ H_2O + NH_3 \rightarrow NH_4^+ + OH^- \\
5. & \ Al(OH)_3 + H_3PO_4 \rightarrow AlPO_4 + 3 H_2O \\
6. & \ 3 H_2S + Al_2(CO_3)_3 \rightarrow Al_2S_3 + 3 CO_2 + 3 H_2O \\
7. & \ MgO + 2 HF \rightarrow MgF_2 + H_2O \\
8. & \ H_2C_2O_4 + Ca(OH)_2 \rightarrow Ca(C_2H_3O_2)_2 + H_2O \\
9. & \ CaCO_3 + 2 HNO_3 \rightarrow Ca(NO_3)_2 + CO_2 + H_2O \\
10. & \ H_2C_2O_4 + 2 LiOH \rightarrow Li_2C_2O_4 + 2 H_2O
\end{aligned}
}
\]