Question

A double replacement chemical equation is shown:
_MnO2 + _HCI → _MnCI2 + _H2O + _CI1
What values should be inserted into the blanks as coefficients to ensure the preservation of the law of conservation of matter?

Answer 1

To balance the chemical equation \( \text{MnO}_2 + \text{HCl} \rightarrow \text{MnCl}_2 + \text{H}_2\text{O} + \text{Cl}_2 \), we need to ensure that the number of each type of atom is the same on both sides of the equation.

Let's break it down step-by-step:

1. **Count the atoms of each element on both sides:**

- Reactants:
- Mn: 1 (from \( \text{MnO}_2 \))
- O: 2 (from \( \text{MnO}_2 \))
- H: 1 (from \( \text{HCl} \))
- Cl: 1 (from \( \text{HCl} \))

- Products:
- Mn: 1 (from \( \text{MnCl}_2 \))
- Cl: 2 (from \( \text{MnCl}_2 \)) + 2 (from \( \text{Cl}_2 \)) = 4 total Cl
- H: 2 (from \( \text{H}_2\text{O} \))
- O: 1 (from \( \text{H}_2\text{O} \))

2. **Start balancing**:

- The manganese (Mn) atoms are balanced (1 on each side).
- We need to balance the chlorine (Cl) and hydrogen (H) atoms. To do this:
- Place a coefficient of 4 in front of \( \text{HCl} \) on the left side to yield 4 Cl and 4 H, and we also keep the \( \text{MnO}_2 \) for oxygen balance.
- We now have:
\[
\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2
\]
- Now go through each element again to confirm balance:
- Mn: 1 on both sides
- O: 2 (from \( \text{MnO}_2 \)) and 2 (from 2 \( \text{H}_2O \))
- H: 4 on the left (from \( 4 \text{HCl} \)) and 4 on the right (from \( 2 \text{H}_2O \))
- Cl: 4 from \( 4 \text{HCl} \) and 2 (from \( \text{MnCl}_2 \)) + 2 (from \( \text{Cl}_2 \)) = 4 on the right side

3. **Final balanced equation**:
\[
1\text{MnO}_2 + 4\text{HCl} \rightarrow 1\text{MnCl}_2 + 2\text{H}_2\text{O} + 1\text{Cl}_2
\]

4. **Insert the coefficients**:
Therefore, the final coefficients for the equation are:
- MnO2: 1
- HCl: 4
- MnCl2: 1
- H2O: 2
- Cl2: 1

So, the coefficients inserted would be:

1, 4, 1, 2, 1.