Answer :
Certainly! Let's balance each chemical equation step by step.
### Equation 1:
[tex]\[ KHF_2 \rightarrow KF + H_2 + F_2 \][/tex]
1. Balancing Potassium (K):
- On the left side, we have 1 atom of K in \(KHF_2\).
- To balance potassium atoms, we need 1 atom of K on the right side, which we get from \(KF\).
- Thus, the coefficient for \(KF\) is 1.
2. Balancing Hydrogen (H):
- On the left side, we have 1 atom of H in \(KHF_2\).
- To balance hydrogen atoms, we need \(H_2\) on the right side. We must multiply it by a factor to match the number of hydrogen atoms.
- To get 2 hydrogen atoms on the right side, the coefficient for \(H_2\) is \(\frac{1}{2}\).
3. Balancing Fluorine (F):
- On the left side, we have 2 atoms of F in \(KHF_2\).
- On the right side, in \(KF\) and \(F_2\), the total number of fluorine atoms needs to be 2. Fluorine atoms in \(KF\) is 1 and \(F_2\) is 2.
- To balance, the coefficient for \(F_2\) is \(\frac{1}{2}\).
Combining the steps, the balanced equation is:
[tex]\[ 1KHF_2 \rightarrow 1KF + \frac{1}{2}H_2 + \frac{1}{2}F_2 \][/tex]
### Equation 2:
[tex]\[ P_4O_{10} + H_2O \longrightarrow H_3PO_4 \][/tex]
1. Balancing Phosphorus (P):
- On the left side, we have 4 phosphorus atoms in \(P_4O_{10}\).
- On the right side, \(H_3PO_4\) contains 1 phosphorus atom per molecule.
- To balance the phosphorus atoms, the coefficient for \(H_3PO_4\) is 4.
2. Balancing Oxygen (O):
- On the left side, we have 10 oxygen atoms in \(P_4O_{10}\) and unknown from \(H_2O\).
- On the right side, using our previous coefficient for \(H_3PO_4\), it contains \(4 \times 4 = 16\) oxygen atoms.
3. Balancing Hydrogen (H):
- On the left side, we have 2 hydrogen atoms in \(H_2O\).
- On the right side, now we have \(4 \times 3 = 12\) hydrogen atoms in \(H_3PO_4\).
Combining the steps, the balanced equation is:
[tex]\[ 1P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4 \][/tex]
### Summary of Balanced Equations:
[tex]\[ 1. \quad KHF_2 \rightarrow KF + \frac{1}{2} H_2 + \frac{1}{2} F_2 \][/tex]
[tex]\[ 2. \quad 1P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4 \][/tex]
### Equation 1:
[tex]\[ KHF_2 \rightarrow KF + H_2 + F_2 \][/tex]
1. Balancing Potassium (K):
- On the left side, we have 1 atom of K in \(KHF_2\).
- To balance potassium atoms, we need 1 atom of K on the right side, which we get from \(KF\).
- Thus, the coefficient for \(KF\) is 1.
2. Balancing Hydrogen (H):
- On the left side, we have 1 atom of H in \(KHF_2\).
- To balance hydrogen atoms, we need \(H_2\) on the right side. We must multiply it by a factor to match the number of hydrogen atoms.
- To get 2 hydrogen atoms on the right side, the coefficient for \(H_2\) is \(\frac{1}{2}\).
3. Balancing Fluorine (F):
- On the left side, we have 2 atoms of F in \(KHF_2\).
- On the right side, in \(KF\) and \(F_2\), the total number of fluorine atoms needs to be 2. Fluorine atoms in \(KF\) is 1 and \(F_2\) is 2.
- To balance, the coefficient for \(F_2\) is \(\frac{1}{2}\).
Combining the steps, the balanced equation is:
[tex]\[ 1KHF_2 \rightarrow 1KF + \frac{1}{2}H_2 + \frac{1}{2}F_2 \][/tex]
### Equation 2:
[tex]\[ P_4O_{10} + H_2O \longrightarrow H_3PO_4 \][/tex]
1. Balancing Phosphorus (P):
- On the left side, we have 4 phosphorus atoms in \(P_4O_{10}\).
- On the right side, \(H_3PO_4\) contains 1 phosphorus atom per molecule.
- To balance the phosphorus atoms, the coefficient for \(H_3PO_4\) is 4.
2. Balancing Oxygen (O):
- On the left side, we have 10 oxygen atoms in \(P_4O_{10}\) and unknown from \(H_2O\).
- On the right side, using our previous coefficient for \(H_3PO_4\), it contains \(4 \times 4 = 16\) oxygen atoms.
3. Balancing Hydrogen (H):
- On the left side, we have 2 hydrogen atoms in \(H_2O\).
- On the right side, now we have \(4 \times 3 = 12\) hydrogen atoms in \(H_3PO_4\).
Combining the steps, the balanced equation is:
[tex]\[ 1P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4 \][/tex]
### Summary of Balanced Equations:
[tex]\[ 1. \quad KHF_2 \rightarrow KF + \frac{1}{2} H_2 + \frac{1}{2} F_2 \][/tex]
[tex]\[ 2. \quad 1P_4O_{10} + 6H_2O \longrightarrow 4H_3PO_4 \][/tex]
