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 + H3PO4 → Ca3(PO4)2 + H2O?

Question

GPT-4o mini · Answer

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 ensure that the number of each type of atom on the reactants side equals the number on the products side. 1. **Write down the unbalanced equation:** $$ \text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} $$ 2. **Identify the number of each type of atom in the reactants and products:** - Reactants: - Calcium (Ca): 1 (from $\text{Ca(OH)}_2$) - Oxygen (O): 2 (from $\text{Ca(OH)}_2$) + 4 (from $\text{H}_3\text{PO}_4$) = 6 - Hydrogen (H): 2 (from $\text{Ca(OH)}_2$) + 3 (from $\text{H}_3\text{PO}_4$) = 5 - Phosphorus (P): 1 (from $\text{H}_3\text{PO}_4$) - Products: - Calcium (Ca): 3 (from $\text{Ca}_3(\text{PO}_4)_2$) - Oxygen (O): 8 (from $\text{Ca}_3(\text{PO}_4)_2$) + 1 (from $\text{H}_2\text{O}$) = 9 - Hydrogen (H): 2 (from $\text{H}_2\text{O}$) - Phosphorus (P): 2 (from $\text{Ca}_3(\text{PO}_4)_2$) 3. **Balance the equation step by step:** - Start by balancing the $\text{Ca}$ atoms. Since there are 3 $\text{Ca}$ in the products, we need 3 $\text{Ca(OH)}_2$ in the reactants: $$ 3\text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} $$ - Now we have: - Reactants: - Ca: 3 - O: 6 (from $3\text{Ca(OH)}_2$) + 4 (from $\text{H}_3\text{PO}_4$) = 10 - H: 6 (from $3\text{Ca(OH)}_2$) + 3 = 9 - P: 1 - Next, we need to balance the phosphorus atoms. We have 2 phosphorus in the products, so we will use 2 $\text{H}_3\text{PO}_4$: $$ 3\text{Ca(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} $$ - Reevaluate the counts: - Reactants: - Ca: 3 - O: 6 (from $3\text{Ca(OH)}_2$) + 8 (from $2\text{H}_3\text{PO}_4$) = 14 - H: 6 (from $3\text{Ca(OH)}_2$) + 6 (from $2\text{H}_3\text{PO}_4$) = 12 - P: 2 - Now we have: - Products: - Ca: 3 - O: 8 (from $\text{Ca}_3(\text{PO}_4)_2$) + 1 (from $\text{H}_2\text{O}$) to be determined - H: 2 (from $\text{H}_2\text{O}$) - P: 2 - Balance hydrogen and oxygen: - We need 10 oxygen atoms on the product side, so since $ \text{H}_2\text{O} $ has 1 oxygen, we will need to add 6 $ \text{H}_2\text{O} $: $$ 3\text{Ca(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O} $$ 4. **Final counts for the balanced equation:** - Reactants: - Ca: 3 - O: 14 - H: 12 - P: 2 - Products: - Ca: 3 - O: 14 (8 from $ \text{Ca}_3(\text{PO}_4)_2 + 6 \text{H}_2\text{O} $) - H: 12 - P: 2 Thus, the balanced equation is: $$ 3\text{Ca(OH)}_2 + 2\text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O} $$