Answer :
To balance the chemical reaction,
[tex]\[ \text { ? } CaCO _3+\text { ? } H _3 PO _4 \longrightarrow \text { ? } Ca _3\left( PO _4\right)_2+\text { ? } H _2 O +\text { ? } CO _2 \][/tex]
we will follow a systematic approach to ensure that we have the smallest possible integer coefficients for each compound, making sure that the number of atoms of each element is conserved in the reaction.
### Step-by-Step Balancing:
1. Identify each element present in the reaction:
- Calcium (Ca)
- Carbon (C)
- Oxygen (O)
- Hydrogen (H)
- Phosphorus (P)
2. Write the unbalanced equation with placeholders for coefficients:
[tex]\[ a \, CaCO_3 + b \, H_3PO_4 \longrightarrow c \, Ca_3(PO_4)_2 + d \, H_2O + e \, CO_2 \][/tex]
3. Balance calcium (Ca):
- Each molecule of [tex]\( Ca_3(PO_4)_2 \)[/tex] will introduce 3 calcium atoms.
- Thus, if we have [tex]\( c \, Ca_3(PO_4)_2 \)[/tex], we need [tex]\( 3c \)[/tex] Ca from [tex]\( CaCO_3 \)[/tex].
[tex]\[ a = 3c \][/tex]
4. Balance phosphorus (P):
- Each molecule of [tex]\( Ca_3(PO_4)_2 \)[/tex] has 2 phosphorus atoms.
- Thus, if we have [tex]\( c \, Ca_3(PO_4)_2 \)[/tex], we need [tex]\( 2c \)[/tex] phosphates from [tex]\( H_3PO_4 \)[/tex].
[tex]\[ b = 2c \][/tex]
5. Balance carbon (C):
- Each molecule of [tex]\( CaCO_3 \)[/tex] provides 1 carbon atom.
- Each molecule of [tex]\( CO_2 \)[/tex] has 1 carbon atom.
[tex]\[ a = e \][/tex]
6. Balance hydrogen (H):
- Each molecule of [tex]\( H_3PO_4 \)[/tex] provides 3 hydrogen atoms.
- Each molecule of [tex]\( H_2O \)[/tex] has 2 hydrogen atoms.
- Consequently,
[tex]\[ 3b = 2d \quad \Rightarrow \quad d = \frac{3b}{2} \][/tex]
7. Balance oxygen (O):
- Total oxygen atoms on the left:
[tex]\[ 3a + 4b \][/tex]
- Total oxygen atoms on the right:
[tex]\[ 8c (from \, Ca_3(PO_4)_2) + d (from \, H_2O) + 2e (from \, CO_2) \][/tex]
- Equating,
[tex]\[ 3a + 4b = 8c + d + 2e \][/tex]
### Solving the Equations:
From the stepwise balancing:
1. [tex]\( a = 3c \)[/tex]
2. [tex]\( b = 2c \)[/tex]
3. [tex]\( a = e \)[/tex] implies [tex]\( 3c = e \)[/tex]
4. Substituting [tex]\( b = 2c \)[/tex] and [tex]\( d = \frac{3b}{2} = \frac{3 \cdot 2c}{2} = 3c \)[/tex]
5. From the oxygen balance: [tex]\( 3a + 4b = 8c + d + 2e \)[/tex]:
Substitute [tex]\( a, b, d, e \)[/tex]:
[tex]\[ 3(3c) + 4(2c) = 8c + 3c + 2(3c) \][/tex]
[tex]\[ 9c + 8c = 8c + 3c + 6c \][/tex]
[tex]\[ 17c = 17c \][/tex]
This shows that the above coefficients satisfy oxygen balance too.
### Solution:
From [tex]\( c = 1 \)[/tex],
[tex]\[ a = 3c = 3 \][/tex]
[tex]\[ b = 2c = 2 \][/tex]
[tex]\[ d = 3c = 3 \][/tex]
[tex]\[ e = 3c = 3 \][/tex]
Thus, the balanced equation is:
[tex]\[ 3 \, CaCO_3 + 2 \, H_3PO_4 \longrightarrow 1 \, Ca_3(PO_4)_2 + 3 \, H_2O + 3 \, CO_2 \][/tex]
The coefficient of carbon dioxide [tex]\( CO_2 \)[/tex] is:
[tex]\[ \boxed{3} \][/tex]
[tex]\[ \text { ? } CaCO _3+\text { ? } H _3 PO _4 \longrightarrow \text { ? } Ca _3\left( PO _4\right)_2+\text { ? } H _2 O +\text { ? } CO _2 \][/tex]
we will follow a systematic approach to ensure that we have the smallest possible integer coefficients for each compound, making sure that the number of atoms of each element is conserved in the reaction.
### Step-by-Step Balancing:
1. Identify each element present in the reaction:
- Calcium (Ca)
- Carbon (C)
- Oxygen (O)
- Hydrogen (H)
- Phosphorus (P)
2. Write the unbalanced equation with placeholders for coefficients:
[tex]\[ a \, CaCO_3 + b \, H_3PO_4 \longrightarrow c \, Ca_3(PO_4)_2 + d \, H_2O + e \, CO_2 \][/tex]
3. Balance calcium (Ca):
- Each molecule of [tex]\( Ca_3(PO_4)_2 \)[/tex] will introduce 3 calcium atoms.
- Thus, if we have [tex]\( c \, Ca_3(PO_4)_2 \)[/tex], we need [tex]\( 3c \)[/tex] Ca from [tex]\( CaCO_3 \)[/tex].
[tex]\[ a = 3c \][/tex]
4. Balance phosphorus (P):
- Each molecule of [tex]\( Ca_3(PO_4)_2 \)[/tex] has 2 phosphorus atoms.
- Thus, if we have [tex]\( c \, Ca_3(PO_4)_2 \)[/tex], we need [tex]\( 2c \)[/tex] phosphates from [tex]\( H_3PO_4 \)[/tex].
[tex]\[ b = 2c \][/tex]
5. Balance carbon (C):
- Each molecule of [tex]\( CaCO_3 \)[/tex] provides 1 carbon atom.
- Each molecule of [tex]\( CO_2 \)[/tex] has 1 carbon atom.
[tex]\[ a = e \][/tex]
6. Balance hydrogen (H):
- Each molecule of [tex]\( H_3PO_4 \)[/tex] provides 3 hydrogen atoms.
- Each molecule of [tex]\( H_2O \)[/tex] has 2 hydrogen atoms.
- Consequently,
[tex]\[ 3b = 2d \quad \Rightarrow \quad d = \frac{3b}{2} \][/tex]
7. Balance oxygen (O):
- Total oxygen atoms on the left:
[tex]\[ 3a + 4b \][/tex]
- Total oxygen atoms on the right:
[tex]\[ 8c (from \, Ca_3(PO_4)_2) + d (from \, H_2O) + 2e (from \, CO_2) \][/tex]
- Equating,
[tex]\[ 3a + 4b = 8c + d + 2e \][/tex]
### Solving the Equations:
From the stepwise balancing:
1. [tex]\( a = 3c \)[/tex]
2. [tex]\( b = 2c \)[/tex]
3. [tex]\( a = e \)[/tex] implies [tex]\( 3c = e \)[/tex]
4. Substituting [tex]\( b = 2c \)[/tex] and [tex]\( d = \frac{3b}{2} = \frac{3 \cdot 2c}{2} = 3c \)[/tex]
5. From the oxygen balance: [tex]\( 3a + 4b = 8c + d + 2e \)[/tex]:
Substitute [tex]\( a, b, d, e \)[/tex]:
[tex]\[ 3(3c) + 4(2c) = 8c + 3c + 2(3c) \][/tex]
[tex]\[ 9c + 8c = 8c + 3c + 6c \][/tex]
[tex]\[ 17c = 17c \][/tex]
This shows that the above coefficients satisfy oxygen balance too.
### Solution:
From [tex]\( c = 1 \)[/tex],
[tex]\[ a = 3c = 3 \][/tex]
[tex]\[ b = 2c = 2 \][/tex]
[tex]\[ d = 3c = 3 \][/tex]
[tex]\[ e = 3c = 3 \][/tex]
Thus, the balanced equation is:
[tex]\[ 3 \, CaCO_3 + 2 \, H_3PO_4 \longrightarrow 1 \, Ca_3(PO_4)_2 + 3 \, H_2O + 3 \, CO_2 \][/tex]
The coefficient of carbon dioxide [tex]\( CO_2 \)[/tex] is:
[tex]\[ \boxed{3} \][/tex]
