Question

Balance the following chemical equations
FeCI3 + NaOH = Fe(OH)3 + NaCI

Answer 1

K3PO4 + HCI = KCI + H3PO4

Answer 2

CaBr2 + Na3P = Ca3P2 + NaBr

Answer 3

B2Br6 + HNO3 = B(NO3)e + HBr

Answer 4

To balance the chemical equation for the reaction between iron(III) chloride (FeCl₃) and sodium hydroxide (NaOH) to form iron(III) hydroxide (Fe(OH)₃) and sodium chloride (NaCl), follow these steps:

1. Write the unbalanced equation:
\[
\text{FeCl}_3 + \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaCl}
\]

2. Identify the number of each type of atom on both sides of the equation:
- **Reactants**:
- Fe: 1 in FeCl₃
- Cl: 3 in FeCl₃
- Na: 1 in NaOH
- O: 1 in NaOH
- H: 1 in NaOH

- **Products**:
- Fe: 1 in Fe(OH)₃
- Cl: 1 in NaCl
- Na: 1 in NaCl
- O: 3 in Fe(OH)₃
- H: 3 in Fe(OH)₃

3. Balance the equation by adjusting the coefficients:
- **Iron (Fe)** is balanced with 1 on each side.
- **Chlorine (Cl)** has 3 on the left (in FeCl₃) and only 1 on the right (in NaCl), so we need 3 NaCl:
\[
\text{FeCl}_3 + \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl}
\]
- Now we have 3 Na in the products, so put a coefficient of 3 for NaOH:
\[
\text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl}
\]

4. Now recount the atoms:
- **Reactants**:
- Fe: 1
- Cl: 3
- Na: 3
- O: 3
- H: 3
- **Products**:
- Fe: 1
- Cl: 3
- Na: 3
- O: 3
- H: 3

Both sides are balanced.

The balanced equation is:
\[
\text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl}
\]

Answer 5

To balance the chemical equation for the reaction between potassium phosphate (K₃PO₄) and hydrochloric acid (HCl) to form potassium chloride (KCl) and phosphoric acid (H₃PO₄), follow these steps:

1. **Write the unbalanced equation**:
\[
\text{K}_3\text{PO}_4 + \text{HCl} \rightarrow \text{KCl} + \text{H}_3\text{PO}_4
\]

2. **Count the number of each atom on both sides**:
- **Reactants**:
- K: 3 (in K₃PO₄)
- P: 1 (in K₃PO₄)
- O: 4 (in K₃PO₄)
- H: 1 (in HCl)
- Cl: 1 (in HCl)

- **Products**:
- K: 1 (in KCl)
- Cl: 1 (in KCl)
- H: 3 (in H₃PO₄)
- P: 1 (in H₃PO₄)
- O: 4 (in H₃PO₄)

3. **Balance the equation**:
- **Potassium (K)**: There are 3 K atoms in the reactants, so we need to put a coefficient of 3 in front of KCl:
\[
\text{K}_3\text{PO}_4 + \text{HCl} \rightarrow 3\text{KCl} + \text{H}_3\text{PO}_4
\]
- Now recount the reactants:
- K: 3 (in K₃PO₄)
- H: 1 (in HCl)
- Cl: 3 (from 3 KCl)

- To balance the 3 HCl needed, we now adjust the coefficient in front of HCl to 3:
\[
\text{K}_3\text{PO}_4 + 3\text{HCl} \rightarrow 3\text{KCl} + \text{H}_3\text{PO}_4
\]

4. **Final count**:
- **Reactants**:
- K: 3
- P: 1
- O: 4
- H: 3
- Cl: 3

- **Products**:
- K: 3
- Cl: 3
- H: 3
- P: 1
- O: 4

Both sides are balanced.

The final balanced equation is:
\[
\text{K}_3\text{PO}_4 + 3\text{HCl} \rightarrow 3\text{KCl} + \text{H}_3\text{PO}_4
\]

Answer 6

To balance the chemical equation for the reaction between calcium bromide (CaBr₂) and sodium phosphate (Na₃P) to form calcium phosphate (Ca₃P₂) and sodium bromide (NaBr), follow these steps:

1. **Write the unbalanced equation**:
\[
\text{CaBr}_2 + \text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + \text{NaBr}
\]

2. **Count the number of each type of atom on both sides**:
- **Reactants**:
- Ca: 1 (in CaBr₂)
- Br: 2 (in CaBr₂)
- Na: 3 (in Na₃P)
- P: 1 (in Na₃P)

- **Products**:
- Ca: 3 (in Ca₃P₂)
- P: 2 (in Ca₃P₂)
- Br: 1 (in NaBr)
- Na: 1 (in NaBr)

3. **Balance the equation**:
- Start by balancing calcium (Ca):
- Since there are 3 Ca in the products (Ca₃P₂), we need 3 CaBr₂ in the reactants:
\[
3\text{CaBr}_2 + \text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + \text{NaBr}
\]

- Now, recount:
- **Reactants**:
- Ca: 3
- Br: 6 (from 3 CaBr₂)
- Na: 3 (from Na₃P)
- P: 1

- **Products**:
- Ca: 3
- P: 2
- Br: 1 (from NaBr)
- Na: 1

- Next, balance phosphorus (P). Since we have 2 P in the products (Ca₃P₂), we need to have 2 Na₃P:
\[
3\text{CaBr}_2 + 2\text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + \text{NaBr}
\]

- Now recount:
- **Reactants**:
- Ca: 3
- Br: 6 (3 CaBr₂)
- Na: 6 (2 Na₃P)
- P: 2 (2 Na₃P)

- **Products**:
- Ca: 3
- P: 2
- Br: 1 (from NaBr)
- Na: 1 (from NaBr)

- Finally, balance sodium (Na):
- Since we have 6 Na in the reactants, we need to have 6 NaBr in the products:
\[
3\text{CaBr}_2 + 2\text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + 6\text{NaBr}
\]

4. **Final count**:
- **Reactants**:
- Ca: 3
- Br: 6
- Na: 6 (from 2 Na₃P)
- P: 2 (from 2 Na₃P)

- **Products**:
- Ca: 3 (from Ca₃P₂)
- P: 2 (from Ca₃P₂)
- Na: 6 (from 6 NaBr)
- Br: 6 (from 6 NaBr)

Both sides are balanced.

The final balanced equation is:
\[
3\text{CaBr}_2 + 2\text{Na}_3\text{P} \rightarrow \text{Ca}_3\text{P}_2 + 6\text{NaBr}
\]

Answer 7

To balance the chemical equation for the reaction between dibromine (B₂Br₆) and nitric acid (HNO₃) to form boron nitrate (B(NO₃)₃) and hydrobromic acid (HBr), we can follow these steps:

1. **Write the unbalanced equation**:
\[
\text{B}_2\text{Br}_6 + \text{HNO}_3 \rightarrow \text{B(NO}_3)_3 + \text{HBr}
\]

2. **Identify and count the number of atoms on both sides**:
- **Reactants**:
- B: 2 (in B₂Br₆)
- Br: 6 (in B₂Br₆)
- H: 1 (in HNO₃)
- N: 1 (in HNO₃)
- O: 3 (in HNO₃)

- **Products**:
- B: 1 (in B(NO₃)₃)
- Br: 1 (in HBr)
- H: 1 (in HBr)
- N: 3 (in B(NO₃)₃, one for each nitrate group)
- O: 9 (in B(NO₃)₃, three nitrate groups each containing three oxygen atoms)

3. **Balance the equation**:
- Start with boron (B):
- Since we have 2 B in the reactants, we need 2 B(NO₃)₃ in the products:
\[
\text{B}_2\text{Br}_6 + \text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + \text{HBr}
\]
- Now recount:
- **Products** now have:
- B: 2 (from 2 B(NO₃)₃)
- N: 6 (from 2 B(NO₃)₃)
- H: 1 (from HBr)
- O: 18 (from 2 B(NO₃)₃)
- Br: 1 (from HBr)
- In order to balance the nitrogen and hydrobromic acid, we need to put in the appropriate coefficients.

4. **Determine how many nitric acid and hydrobromic acid are needed**:
- There are 6 nitrogen atoms in total from 2 B(NO₃)₃. Thus, we need 6 HNO₃:
\[
\text{B}_2\text{Br}_6 + 6\text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + \text{HBr}
\]

- For each bromine atom in B₂Br₆, we need a corresponding HBr, which means we need 6 HBr to balance the 6 Br:
\[
\text{B}_2\text{Br}_6 + 6\text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + 6\text{HBr}
\]

5. **Final count**:
- **Reactants**:
- B: 2 (from B₂Br₆)
- Br: 6 (from B₂Br₆)
- H: 6 (from 6 HNO₃)
- N: 6 (from 6 HNO₃)
- O: 18 (from 6 HNO₃)

- **Products**:
- B: 2 (from 2 B(NO₃)₃)
- Br: 6 (from 6 HBr)
- H: 6 (from 6 HBr)
- N: 6 (from 2 B(NO₃)₃)
- O: 18 (from 2 B(NO₃)₃)

Now both sides are balanced.

The final balanced equation is:
\[
\text{B}_2\text{Br}_6 + 6\text{HNO}_3 \rightarrow 2\text{B(NO}_3)_3 + 6\text{HBr}
\]