Answer :
To balance the chemical equation given:
[tex]\[ Ca(OH)_2 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
we need to ensure that the number of atoms of each element on the reactant side equals the number of atoms on the product side. Let's go through the balancing process step-by-step.
### Step 1: Balance the Calcium (Ca) Atoms
First, observe the calcium atoms. There are 3 calcium atoms on the right side in [tex]\( Ca_3(PO_4)_2 \)[/tex]. Therefore, we need 3 calcium atoms on the left side. We can achieve this by placing a coefficient of 3 in front of [tex]\( Ca(OH)_2 \)[/tex]:
[tex]\[ 3Ca(OH)_2 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
### Step 2: Balance the Phosphate (PO_4) Groups
Next, look at the phosphate (PO_4) groups. There are 2 phosphate groups on the right side in [tex]\( Ca_3(PO_4)_2 \)[/tex]. Therefore, we need 2 phosphate groups on the left side. We can achieve this by placing a coefficient of 2 in front of [tex]\( H_3PO_4 \)[/tex]:
[tex]\[ 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
### Step 3: Balance the Hydrogen (H) Atoms
Now consider the hydrogen atoms. On the left side, there are:
- 3 molecules of [tex]\( Ca(OH)_2 \)[/tex] each contributing 2 hydrogen atoms, giving a total of [tex]\( 3 \times 2 = 6 \)[/tex] hydrogen atoms,
- Plus 2 molecules of [tex]\( H_3PO_4 \)[/tex] each contributing 3 hydrogen atoms, giving a total of [tex]\( 2 \times 3 = 6 \)[/tex] hydrogen atoms.
This sums to [tex]\( 6 + 6 = 12 \)[/tex] hydrogen atoms on the left side.
On the right side, we have hydrogen atoms in water ([tex]\( H_2O \)[/tex]). Since each water molecule contains 2 hydrogen atoms, we need 6 water molecules (each contributing 2 hydrogen atoms) to get a total of 12 hydrogen atoms:
[tex]\[ 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O \][/tex]
### Step 4: Balance the Oxygen (O) Atoms
Finally, check the oxygen atoms. On the left side, we have:
- 3 molecules of [tex]\( Ca(OH)_2 \)[/tex] each contributing 2 oxygen atoms, giving a total of [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms,
- Plus 2 molecules of [tex]\( H_3PO_4 \)[/tex] each contributing 4 oxygen atoms, giving a total of [tex]\( 2 \times 4 = 8 \)[/tex] oxygen atoms.
This sums to [tex]\( 6 + 8 = 14 \)[/tex] oxygen atoms on the left side.
On the right side, we have:
- 2 phosphate groups in [tex]\( Ca_3(PO_4)_2 \)[/tex], each phosphate group contributing 4 oxygen atoms, giving a total of [tex]\( 2 \times 4 = 8 \)[/tex] oxygen atoms,
- Plus 6 water molecules, each contributing 1 oxygen atom, giving a total of [tex]\( 6 \times 1 = 6 \)[/tex] oxygen atoms.
This also sums to [tex]\( 8 + 6 = 14 \)[/tex] oxygen atoms on the right side.
Since all elements are now balanced, the balanced chemical equation is:
[tex]\[ 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O \][/tex]
The coefficient that comes in front of [tex]\( Ca(OH)_2 \)[/tex] is [tex]\( 3 \)[/tex].
[tex]\[ Ca(OH)_2 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
we need to ensure that the number of atoms of each element on the reactant side equals the number of atoms on the product side. Let's go through the balancing process step-by-step.
### Step 1: Balance the Calcium (Ca) Atoms
First, observe the calcium atoms. There are 3 calcium atoms on the right side in [tex]\( Ca_3(PO_4)_2 \)[/tex]. Therefore, we need 3 calcium atoms on the left side. We can achieve this by placing a coefficient of 3 in front of [tex]\( Ca(OH)_2 \)[/tex]:
[tex]\[ 3Ca(OH)_2 + H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
### Step 2: Balance the Phosphate (PO_4) Groups
Next, look at the phosphate (PO_4) groups. There are 2 phosphate groups on the right side in [tex]\( Ca_3(PO_4)_2 \)[/tex]. Therefore, we need 2 phosphate groups on the left side. We can achieve this by placing a coefficient of 2 in front of [tex]\( H_3PO_4 \)[/tex]:
[tex]\[ 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + H_2O \][/tex]
### Step 3: Balance the Hydrogen (H) Atoms
Now consider the hydrogen atoms. On the left side, there are:
- 3 molecules of [tex]\( Ca(OH)_2 \)[/tex] each contributing 2 hydrogen atoms, giving a total of [tex]\( 3 \times 2 = 6 \)[/tex] hydrogen atoms,
- Plus 2 molecules of [tex]\( H_3PO_4 \)[/tex] each contributing 3 hydrogen atoms, giving a total of [tex]\( 2 \times 3 = 6 \)[/tex] hydrogen atoms.
This sums to [tex]\( 6 + 6 = 12 \)[/tex] hydrogen atoms on the left side.
On the right side, we have hydrogen atoms in water ([tex]\( H_2O \)[/tex]). Since each water molecule contains 2 hydrogen atoms, we need 6 water molecules (each contributing 2 hydrogen atoms) to get a total of 12 hydrogen atoms:
[tex]\[ 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O \][/tex]
### Step 4: Balance the Oxygen (O) Atoms
Finally, check the oxygen atoms. On the left side, we have:
- 3 molecules of [tex]\( Ca(OH)_2 \)[/tex] each contributing 2 oxygen atoms, giving a total of [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms,
- Plus 2 molecules of [tex]\( H_3PO_4 \)[/tex] each contributing 4 oxygen atoms, giving a total of [tex]\( 2 \times 4 = 8 \)[/tex] oxygen atoms.
This sums to [tex]\( 6 + 8 = 14 \)[/tex] oxygen atoms on the left side.
On the right side, we have:
- 2 phosphate groups in [tex]\( Ca_3(PO_4)_2 \)[/tex], each phosphate group contributing 4 oxygen atoms, giving a total of [tex]\( 2 \times 4 = 8 \)[/tex] oxygen atoms,
- Plus 6 water molecules, each contributing 1 oxygen atom, giving a total of [tex]\( 6 \times 1 = 6 \)[/tex] oxygen atoms.
This also sums to [tex]\( 8 + 6 = 14 \)[/tex] oxygen atoms on the right side.
Since all elements are now balanced, the balanced chemical equation is:
[tex]\[ 3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O \][/tex]
The coefficient that comes in front of [tex]\( Ca(OH)_2 \)[/tex] is [tex]\( 3 \)[/tex].
