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Sagot :
To balance the chemical equation [tex]\( KClO_3 \rightarrow KCl + O_2 \)[/tex], we need to determine the correct coefficients that ensure the same number of each type of atom on both sides of the reaction.
Let's follow these steps to balance the equation:
1. Identify the elements involved:
- Potassium (K)
- Chlorine (Cl)
- Oxygen (O)
2. Write the unbalanced equation:
[tex]\[ KClO_3 \rightarrow KCl + O_2 \][/tex]
3. Balance the potassium (K) and chlorine (Cl) atoms:
- There is 1 K on the left side and 1 K on the right side.
- There is 1 Cl on the left side and 1 Cl on the right side.
- Therefore, initially, no extra coefficients are needed for K and Cl at this stage. However, we will adjust this after balancing oxygen.
4. Balance the oxygen (O) atoms:
- On the left side, there are 3 oxygen atoms in [tex]\( KClO_3 \)[/tex].
- On the right side, oxygen is in [tex]\( O_2 \)[/tex], which means it appears in pairs. We need to find a coefficient that gives a total count of oxygen atoms that matches the left side.
5. To balance oxygen, we can put a coefficient of 2 in front of [tex]\( KClO_3 \)[/tex], which gives us:
[tex]\[ 2 KClO_3 \rightarrow KCl + O_2 \][/tex]
6. Adjust the left side and check if it affects other atoms:
- Now, we have 2 K and 2 Cl on the left side because of the coefficient 2 in front of [tex]\( KClO_3 \)[/tex].
To balance the K and Cl atoms:
- We should put a coefficient of 2 in front of [tex]\( KCl \)[/tex], making sure we have 2 K and 2 Cl on the right side:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + O_2 \][/tex]
7. Now let’s balance the oxygen atoms by determining the coefficient for [tex]\( O_2 \)[/tex]:
- On the left side, we have [tex]\( 2 \times 3 = 6 \)[/tex] oxygen atoms (from [tex]\( 2 KClO_3 \)[/tex]).
- On the right side, we currently have 2 oxygen atoms.
- To balance this, we need a coefficient of 3 in front of [tex]\( O_2 \)[/tex] because [tex]\( 3 \times 2 = 6 \)[/tex].
Hence, the fully balanced equation with the correct coefficients is:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]
So, the balanced equation is:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]
Let's follow these steps to balance the equation:
1. Identify the elements involved:
- Potassium (K)
- Chlorine (Cl)
- Oxygen (O)
2. Write the unbalanced equation:
[tex]\[ KClO_3 \rightarrow KCl + O_2 \][/tex]
3. Balance the potassium (K) and chlorine (Cl) atoms:
- There is 1 K on the left side and 1 K on the right side.
- There is 1 Cl on the left side and 1 Cl on the right side.
- Therefore, initially, no extra coefficients are needed for K and Cl at this stage. However, we will adjust this after balancing oxygen.
4. Balance the oxygen (O) atoms:
- On the left side, there are 3 oxygen atoms in [tex]\( KClO_3 \)[/tex].
- On the right side, oxygen is in [tex]\( O_2 \)[/tex], which means it appears in pairs. We need to find a coefficient that gives a total count of oxygen atoms that matches the left side.
5. To balance oxygen, we can put a coefficient of 2 in front of [tex]\( KClO_3 \)[/tex], which gives us:
[tex]\[ 2 KClO_3 \rightarrow KCl + O_2 \][/tex]
6. Adjust the left side and check if it affects other atoms:
- Now, we have 2 K and 2 Cl on the left side because of the coefficient 2 in front of [tex]\( KClO_3 \)[/tex].
To balance the K and Cl atoms:
- We should put a coefficient of 2 in front of [tex]\( KCl \)[/tex], making sure we have 2 K and 2 Cl on the right side:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + O_2 \][/tex]
7. Now let’s balance the oxygen atoms by determining the coefficient for [tex]\( O_2 \)[/tex]:
- On the left side, we have [tex]\( 2 \times 3 = 6 \)[/tex] oxygen atoms (from [tex]\( 2 KClO_3 \)[/tex]).
- On the right side, we currently have 2 oxygen atoms.
- To balance this, we need a coefficient of 3 in front of [tex]\( O_2 \)[/tex] because [tex]\( 3 \times 2 = 6 \)[/tex].
Hence, the fully balanced equation with the correct coefficients is:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]
So, the balanced equation is:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]
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