Bot GPT 4-o mini To determine the mass of calcium hydroxide (\( \text{Ca(OH)}_2 \)) required to precipitate 400 grams of calcium carbonate (\( \text{CaCO}_3 \)), we first need to write the balanced chemical equation for the reaction.
The reaction between calcium hydroxide and carbon dioxide results in the formation of calcium carbonate:
\[ \text{Ca(OH)}_2 (aq) + \text{CO}_2 (g) \rightarrow \text{CaCO}_3 (s) + \text{H}_2\text{O} (l) \]
From this balanced equation, we can see that 1 mole of calcium hydroxide reacts with 1 mole of carbon dioxide to produce 1 mole of calcium carbonate.
Step 1: Calculate the molar mass of \( \text{CaCO}_3 \)
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (3 O atoms)
\[ \text{Molar mass of } \text{CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00) = 100.09 , \text{g/mol} \]
Step 2: Calculate the number of moles of \( \text{CaCO}_3 \) in 400 grams
\[ \text{Number of moles of } \text{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{400 , \text{g}}{100.09 , \text{g/mol}} \approx 3.99 , \text{mol} \]
Step 3: Calculate the number of moles of \( \text{Ca(OH)}_2 \) required
From the balanced equation, we see that the mole ratio between \( \text{Ca(OH)}_2 \) and \( \text{CaCO}_3 \) is 1:1. Therefore, the number of moles of \( \text{Ca(OH)}_2 \) required is also approximately 3.99 mol.
Step 4: Calculate the molar mass of \( \text{Ca(OH)}_2 \)
- Calcium (Ca): 40.08 g/mol
- Oxygen (O): 16.00 g/mol (2 O atoms)
- Hydrogen (H): 1.01 g/mol (2 H atoms)
\[ \text{Molar mass of } \text{Ca(OH)}_2 = 40.08 + (2 \times 16.00) + (2 \times 1.01) = 74.10 , \text{g/mol} \]
Step 5: Calculate the mass of \( \text{Ca(OH)}_2 \) required
\[ \text{Mass of } \text{Ca(OH)}_2 = \text{number of moles} \times \text{molar mass} = 3.99 , \text{mol} \times 74.10 , \text{g/mol} \approx 295.34 , \text{g} \]
Conclusion
To precipitate 400 grams of calcium carbonate (\( \text{CaCO}_3 \)), you would need approximately 295.34 grams of calcium hydroxide (\( \text{Ca(OH)}_2 \)).
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