To calculate the mass of solid calcium carbonate (CaCO₃) that would precipitate when carbon dioxide (CO₂) is passed through a solution of calcium hydroxide (Ca(OH)₂), we can follow these steps:
-
Write the reaction: The relevant chemical reaction is: \[ \text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O} \] Calcium carbonate precipitates out as a solid.
-
Molar mass calculation:
- The molar mass of carbon dioxide (CO₂) is roughly:
\[ 12.01 , (\text{C}) + 16.00 \times 2 , (\text{O}) = 44.01 , \text{g/mol} \] - The molar mass of calcium carbonate (CaCO₃) is roughly:
\[ 40.08 , (\text{Ca}) + 12.01 , (\text{C}) + 16.00 \times 3 , (\text{O}) = 100.09 , \text{g/mol} \]
- The molar mass of carbon dioxide (CO₂) is roughly:
-
Calculate moles of CO₂: \[ \text{Moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{35 , \text{g}}{44.01 , \text{g/mol}} \approx 0.795 , \text{mol} \]
-
Determine moles of CaCO₃ produced: From the balanced chemical equation, 1 mole of CO₂ reacts with 1 mole of Ca(OH)₂ to produce 1 mole of CaCO₃. Therefore, the moles of CaCO₃ produced will be equal to the moles of CO₂ consumed, which is approximately 0.795 mol.
-
Calculate the mass of CaCO₃: \[ \text{Mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 0.795 , \text{mol} \times 100.09 , \text{g/mol} \approx 79.73 , \text{g} \]
Thus, the mass of solid calcium carbonate that would precipitate is approximately 79.73 grams.