Question

Determine the Mass of calcium hydroxide required to precipitate 400grams of caco3 with a equation

Answer 1

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 (\( \text{Ca(OH)}_2 \)).