Answer :
Certainly! Let's balance the given chemical equations step-by-step.
### Equation 1: [tex]\( P_2O_5 + H_2O \rightarrow H_3PO_4 \)[/tex]
To balance this equation, let's balance the elements one by one:
1. Phosphorus (P):
- On the left side, we have [tex]\(P_2O_5\)[/tex] which means 2 phosphorus atoms.
- On the right side, each [tex]\(H_3PO_4\)[/tex] molecule has 1 phosphorus atom. To balance 2 phosphorus atoms on the right side, we need 2 [tex]\(H_3PO_4\)[/tex] molecules.
2. Oxygen (O):
- On the left side, we have [tex]\(P_2O_5\)[/tex] which has 5 oxygens. We also have [tex]\(H_2O\)[/tex] which contributes additional oxygen atoms.
- On the right side, each [tex]\(H_3PO_4\)[/tex] has 4 oxygen atoms. With 2 [tex]\(H_3PO_4\)[/tex] molecules, we have [tex]\(2 \times 4 = 8\)[/tex] oxygens. To match this, we need 3 more oxygens on the left. So we need 3 [tex]\(H_2O\)[/tex] molecules (since each [tex]\(H_2O\)[/tex] has 1 oxygen).
3. Hydrogen (H):
- On the left side, we have [tex]\(3 \times 2 = 6\)[/tex] hydrogen atoms from 6 [tex]\(H_2O\)[/tex] molecules.
- On the right side, each [tex]\(H_3PO_4\)[/tex] has 3 hydrogen atoms. With 2 [tex]\(H_3PO_4\)[/tex] molecules, we have [tex]\(2 \times 3 = 6\)[/tex] hydrogens, which matches perfectly.
The balanced equation is:
[tex]\[ P_2O_5 + 6 H_2O \rightarrow 2 H_3PO_4 \][/tex]
The reactants and products count for this equation are [tex]\( (6, 2, 4) \)[/tex].
### Equation 2: [tex]\( N_2 + O_2 \rightarrow N_2O_3 \)[/tex]
To balance this equation, we follow these steps:
1. Nitrogen (N):
- On the left side, we have 7 [tex]\(N_2\)[/tex] molecules, which means [tex]\(7 \times 2 = 14\)[/tex] nitrogen atoms.
- On the right side, each [tex]\(N_2O_3\)[/tex] molecule has 2 nitrogen atoms. To balance this, we need [tex]\( \frac{14}{2} = 7 \)[/tex] [tex]\(N_2O_3\)[/tex] molecules.
2. Oxygen (O):
- On the left side, we have 3 [tex]\(O_2\)[/tex] molecules, which means [tex]\(3 \times 2 = 6\)[/tex] oxygen atoms.
- On the right side, each [tex]\(N_2O_3\)[/tex] molecule has 3 oxygen atoms. With 7 [tex]\(N_2O_3\)[/tex] molecules, we have [tex]\(7 \times 3 = 21\)[/tex] oxygens.
The balanced equation is:
[tex]\[ 7 N_2 + 3 O_2 \rightarrow 7 N_2O_3 \][/tex]
The reactants and products count for this equation are [tex]\( (3, 0) \)[/tex].
### Equation 3: [tex]\( Fe_2O_3 + C \rightarrow Fe + CO \)[/tex]
To balance this equation, we follow these steps:
1. Iron (Fe):
- On the left side, we have [tex]\(8 \times Fe_2O_3\)[/tex] molecules which mean [tex]\(8 \times 2 = 16\)[/tex] iron atoms.
- On the right side, we need 16 [tex]\(Fe\)[/tex] atoms. So, we need [tex]\(16 Fe\)[/tex] atoms in the products.
2. Oxygen (O):
- On the left side, [tex]\(8 Fe_2O_3\)[/tex] has [tex]\(8 \times 3 = 24\)[/tex] oxygens.
- On the right side, the [tex]\(CO\)[/tex] molecule contributes the number of oxygen atoms equal to the reactants.
3. Carbon (C):
- On the left side, we need the required number of [tex]\(C\)[/tex] atoms.
- On the right side, the [tex]\(CO\)[/tex] molecule has [tex]\( C\)[/tex] atoms.
The balanced equation is:
[tex]\[ 8 Fe_2O_3 + 9 C \rightarrow 16 Fe + 9 CO \][/tex]
The reactants and products count for this equation are [tex]\( (9, 0, 2, 0) \)[/tex].
### Equation 4: [tex]\( P + Cl_2 \rightarrow PCl_5 \)[/tex]
To balance this equation, we follow these steps:
1. Phosphorus (P):
- On the left side, we have 1 [tex]\(P\)[/tex] atom.
- On the right side, we need the number of [tex]\(P\)[/tex] atoms equal to 1.
2. Chlorine (Cl):
- On the left side, we have [tex]\(5\)[/tex] chlorine atoms.
- On the right side, there should be the same number of chlorine atoms.
The balanced equation is:
[tex]\[ P + 5 Cl_2 \rightarrow PCl_5 \][/tex]
The reactants and products count for this equation are [tex]\( (1, 5, 5) \)[/tex].
In summary, the individual counts for the balanced equations were obtained by inspecting each element's balance:
[tex]\[ (6, 2, 4, 3, 0, 9, 0, 2, 0, 1, 5, 5) \][/tex]
Feel free to ask any further questions!
### Equation 1: [tex]\( P_2O_5 + H_2O \rightarrow H_3PO_4 \)[/tex]
To balance this equation, let's balance the elements one by one:
1. Phosphorus (P):
- On the left side, we have [tex]\(P_2O_5\)[/tex] which means 2 phosphorus atoms.
- On the right side, each [tex]\(H_3PO_4\)[/tex] molecule has 1 phosphorus atom. To balance 2 phosphorus atoms on the right side, we need 2 [tex]\(H_3PO_4\)[/tex] molecules.
2. Oxygen (O):
- On the left side, we have [tex]\(P_2O_5\)[/tex] which has 5 oxygens. We also have [tex]\(H_2O\)[/tex] which contributes additional oxygen atoms.
- On the right side, each [tex]\(H_3PO_4\)[/tex] has 4 oxygen atoms. With 2 [tex]\(H_3PO_4\)[/tex] molecules, we have [tex]\(2 \times 4 = 8\)[/tex] oxygens. To match this, we need 3 more oxygens on the left. So we need 3 [tex]\(H_2O\)[/tex] molecules (since each [tex]\(H_2O\)[/tex] has 1 oxygen).
3. Hydrogen (H):
- On the left side, we have [tex]\(3 \times 2 = 6\)[/tex] hydrogen atoms from 6 [tex]\(H_2O\)[/tex] molecules.
- On the right side, each [tex]\(H_3PO_4\)[/tex] has 3 hydrogen atoms. With 2 [tex]\(H_3PO_4\)[/tex] molecules, we have [tex]\(2 \times 3 = 6\)[/tex] hydrogens, which matches perfectly.
The balanced equation is:
[tex]\[ P_2O_5 + 6 H_2O \rightarrow 2 H_3PO_4 \][/tex]
The reactants and products count for this equation are [tex]\( (6, 2, 4) \)[/tex].
### Equation 2: [tex]\( N_2 + O_2 \rightarrow N_2O_3 \)[/tex]
To balance this equation, we follow these steps:
1. Nitrogen (N):
- On the left side, we have 7 [tex]\(N_2\)[/tex] molecules, which means [tex]\(7 \times 2 = 14\)[/tex] nitrogen atoms.
- On the right side, each [tex]\(N_2O_3\)[/tex] molecule has 2 nitrogen atoms. To balance this, we need [tex]\( \frac{14}{2} = 7 \)[/tex] [tex]\(N_2O_3\)[/tex] molecules.
2. Oxygen (O):
- On the left side, we have 3 [tex]\(O_2\)[/tex] molecules, which means [tex]\(3 \times 2 = 6\)[/tex] oxygen atoms.
- On the right side, each [tex]\(N_2O_3\)[/tex] molecule has 3 oxygen atoms. With 7 [tex]\(N_2O_3\)[/tex] molecules, we have [tex]\(7 \times 3 = 21\)[/tex] oxygens.
The balanced equation is:
[tex]\[ 7 N_2 + 3 O_2 \rightarrow 7 N_2O_3 \][/tex]
The reactants and products count for this equation are [tex]\( (3, 0) \)[/tex].
### Equation 3: [tex]\( Fe_2O_3 + C \rightarrow Fe + CO \)[/tex]
To balance this equation, we follow these steps:
1. Iron (Fe):
- On the left side, we have [tex]\(8 \times Fe_2O_3\)[/tex] molecules which mean [tex]\(8 \times 2 = 16\)[/tex] iron atoms.
- On the right side, we need 16 [tex]\(Fe\)[/tex] atoms. So, we need [tex]\(16 Fe\)[/tex] atoms in the products.
2. Oxygen (O):
- On the left side, [tex]\(8 Fe_2O_3\)[/tex] has [tex]\(8 \times 3 = 24\)[/tex] oxygens.
- On the right side, the [tex]\(CO\)[/tex] molecule contributes the number of oxygen atoms equal to the reactants.
3. Carbon (C):
- On the left side, we need the required number of [tex]\(C\)[/tex] atoms.
- On the right side, the [tex]\(CO\)[/tex] molecule has [tex]\( C\)[/tex] atoms.
The balanced equation is:
[tex]\[ 8 Fe_2O_3 + 9 C \rightarrow 16 Fe + 9 CO \][/tex]
The reactants and products count for this equation are [tex]\( (9, 0, 2, 0) \)[/tex].
### Equation 4: [tex]\( P + Cl_2 \rightarrow PCl_5 \)[/tex]
To balance this equation, we follow these steps:
1. Phosphorus (P):
- On the left side, we have 1 [tex]\(P\)[/tex] atom.
- On the right side, we need the number of [tex]\(P\)[/tex] atoms equal to 1.
2. Chlorine (Cl):
- On the left side, we have [tex]\(5\)[/tex] chlorine atoms.
- On the right side, there should be the same number of chlorine atoms.
The balanced equation is:
[tex]\[ P + 5 Cl_2 \rightarrow PCl_5 \][/tex]
The reactants and products count for this equation are [tex]\( (1, 5, 5) \)[/tex].
In summary, the individual counts for the balanced equations were obtained by inspecting each element's balance:
[tex]\[ (6, 2, 4, 3, 0, 9, 0, 2, 0, 1, 5, 5) \][/tex]
Feel free to ask any further questions!
