To determine the number of moles of calcium chloride (CaCl₂) that can be obtained from 25 grams of limestone (calcium carbonate, CaCO₃) in the presence of excess hydrochloric acid (HCl), we can follow these steps:
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Write the balanced chemical reactions:
- The reaction of calcium carbonate (CaCO₃) with hydrochloric acid (HCl) is as follows:
\[ \text{CaCO}_3 (s) + 2\text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]
From this equation, we see that 1 mole of CaCO₃ produces 1 mole of CaCl₂.
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Calculate the molar mass of CaCO₃:
- The molar mass of calcium carbonate (CaCO₃) is calculated by adding the molar masses of each element:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
\[ \text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol} \]
- The molar mass of calcium carbonate (CaCO₃) is calculated by adding the molar masses of each element:
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Calculate the number of moles of CaCO₃ in 25 grams: \[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{25 \text{ g}}{100.09 \text{ g/mol}} \approx 0.2498 \text{ moles} \]
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Determine the number of moles of CaCl₂ produced: According to the balanced equation, 1 mole of CaCO₃ produces 1 mole of CaCl₂. Therefore, the moles of CaCl₂ produced will be the same as the moles of CaCO₃ reacted.
\[ \text{Moles of CaCl}_2 = \text{Moles of CaCO}_3 = 0.2498 \text{ moles} \]
Therefore, from 25 grams of limestone (CaCO₃), approximately 0.2498 moles of calcium chloride (CaCl₂) can be obtained in the presence of excess hydrochloric acid (HCl).