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:
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Count the atoms of each element on both sides:
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Reactants:
- Mn: 1 (from \( \text{MnO}_2 \))
- O: 2 (from \( \text{MnO}_2 \))
- H: 1 (from \( \text{HCl} \))
- Cl: 1 (from \( \text{HCl} \))
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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 \( \text{H}_2\text{O} \))
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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
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Final balanced equation: \[ 1\text{MnO}_2 + 4\text{HCl} \rightarrow 1\text{MnCl}_2 + 2\text{H}_2\text{O} + 1\text{Cl}_2 \]
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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.