Question
Lars is balancing equations with his study group. He is unsure about one equation because each member of the study group came up with a different answer.
Which is the proper way to balance the equation Ca(OH)
2
+H
3
PO
4
→Ca
3
(PO
4
)
2
+H
2
O?
A. Ca(OH)
2
+H
3
PO
4
→Ca
3
(PO
4
)
2
+6H
2
O
B. 3Ca(OH)
2
+2H
3
PO
4
→Ca
3
(PO
4
)
2
+6H
2
O
C. 3Ca(OH)
2
+2H
3
PO
4
→Ca
3
(PO
4
)
2
+3H
2
O
D. Ca(OH)
2
+2H
3
PO
4
→Ca
3
(PO
4
)
2
+6H
2
O
Which is the proper way to balance the equation Ca(OH)
2
+H
3
PO
4
→Ca
3
(PO
4
)
2
+H
2
O?
A. Ca(OH)
2
+H
3
PO
4
→Ca
3
(PO
4
)
2
+6H
2
O
B. 3Ca(OH)
2
+2H
3
PO
4
→Ca
3
(PO
4
)
2
+6H
2
O
C. 3Ca(OH)
2
+2H
3
PO
4
→Ca
3
(PO
4
)
2
+3H
2
O
D. Ca(OH)
2
+2H
3
PO
4
→Ca
3
(PO
4
)
2
+6H
2
O
Answers
Answered by
GPT-4o mini
To balance the chemical equation \( \text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \), we need to count the number of each type of atom on both sides of the equation and make sure they are equal.
1. **Reactants:**
- From \( \text{Ca(OH)}_2 \):
- Calcium (Ca): 1
- Oxygen (O): 2 (from hydroxide)
- Hydrogen (H): 2 (from hydroxide)
- From \( \text{H}_3\text{PO}_4 \):
- Hydrogen (H): 3
- Phosphorus (P): 1
- Oxygen (O): 4
- Total for reactants:
- Ca: 1
- H: 2 + 3 = 5
- P: 1
- O: 2 + 4 = 6
2. **Products:**
- From \( \text{Ca}_3(\text{PO}_4)_2 \):
- Calcium (Ca): 3
- Phosphorus (P): 2 (2 phosphates)
- Oxygen (O): 8 (from phosphates, 4 per phosphate)
- From \( \text{H}_2\text{O} \):
- Hydrogen (H): 2
- Oxygen (O): 1
- Total for products:
- Ca: 3
- H: 2
- P: 2
- O: 8 + 1 = 9
3. **Balancing the equation:**
- To balance Calcium, the easiest step is to multiply \( \text{Ca(OH)}_2 \) by 3, since we need 3 Calcium atoms on the reactant side to match the 3 from the product side.
- To balance Hydrogen, we can keep track of how many we have again after adjusting \( \text{H}_3\text{PO}_4 \) as well, since it contains three Hydrogen atoms, each bringing its total effect to 6 when we need to derive ending water molecules.
Therefore, the correct way to balance this equation is represented by:
**B. \( 3\text{Ca(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O} \)**
This gives us:
- 3 Calcium atoms, 6 Hydrogens (3 from \( \text{H}_3\text{PO}_4 \) and 2 from each water molecule), reconciled to phosphates as needed for the balancing, resulting in a coherent balanced equation.
1. **Reactants:**
- From \( \text{Ca(OH)}_2 \):
- Calcium (Ca): 1
- Oxygen (O): 2 (from hydroxide)
- Hydrogen (H): 2 (from hydroxide)
- From \( \text{H}_3\text{PO}_4 \):
- Hydrogen (H): 3
- Phosphorus (P): 1
- Oxygen (O): 4
- Total for reactants:
- Ca: 1
- H: 2 + 3 = 5
- P: 1
- O: 2 + 4 = 6
2. **Products:**
- From \( \text{Ca}_3(\text{PO}_4)_2 \):
- Calcium (Ca): 3
- Phosphorus (P): 2 (2 phosphates)
- Oxygen (O): 8 (from phosphates, 4 per phosphate)
- From \( \text{H}_2\text{O} \):
- Hydrogen (H): 2
- Oxygen (O): 1
- Total for products:
- Ca: 3
- H: 2
- P: 2
- O: 8 + 1 = 9
3. **Balancing the equation:**
- To balance Calcium, the easiest step is to multiply \( \text{Ca(OH)}_2 \) by 3, since we need 3 Calcium atoms on the reactant side to match the 3 from the product side.
- To balance Hydrogen, we can keep track of how many we have again after adjusting \( \text{H}_3\text{PO}_4 \) as well, since it contains three Hydrogen atoms, each bringing its total effect to 6 when we need to derive ending water molecules.
Therefore, the correct way to balance this equation is represented by:
**B. \( 3\text{Ca(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O} \)**
This gives us:
- 3 Calcium atoms, 6 Hydrogens (3 from \( \text{H}_3\text{PO}_4 \) and 2 from each water molecule), reconciled to phosphates as needed for the balancing, resulting in a coherent balanced equation.
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