To determine how much calcium oxide is produced from the heating of limestone (calcium carbonate), we first need to look at the balanced chemical equation for the decomposition of calcium carbonate:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
From this equation, we can see that 1 mole of calcium carbonate decomposes to produce 1 mole of calcium oxide and 1 mole of carbon dioxide.
-
Calculate the molar masses:
- The molar mass of calcium carbonate \(\text{CaCO}_3\) is:
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol
- Total = 40.08 + 12.01 + 48.00 = 100.09 g/mol (approximately 100 g/mol)
- The molar mass of calcium oxide \(\text{CaO}\) is:
- Ca: 40.08 g/mol
- O: 16.00 g/mol
- Total = 40.08 + 16.00 = 56.08 g/mol (approximately 56 g/mol)
- The molar mass of carbon dioxide \(\text{CO}_2\) is:
- C: 12.01 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
- Total = 12.01 + 32.00 = 44.01 g/mol (approximately 44 g/mol)
- The molar mass of calcium carbonate \(\text{CaCO}_3\) is:
-
Use the stoichiometry of the reaction to find the amounts:
- From the balanced equation: \(1 , \text{mol CaCO}_3\) produces \(1 , \text{mol CaO}\) and \(1 , \text{mol CO}_2\).
- If we start with 200 g of limestone \(\text{CaCO}_3\), then: \[ \text{Moles of } \text{CaCO}_3 = \frac{200 , \text{g}}{100 , \text{g/mol}} = 2 , \text{mol} \]
- Each mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\), therefore: \[ 2 , \text{mol CaCO}_3 \text{ produces } 2 , \text{mol CO}_2 \]
- The mass of carbon dioxide produced is: \[ \text{Mass of } \text{CO}_2 = 2 , \text{mol} \times 44 , \text{g/mol} = 88 , \text{g} \]
- Similarly, the mass of calcium oxide produced will also be: \[ 2 , \text{mol CaCO}_3 \text{ produces } 2 , \text{mol CaO} \] \[ \text{Mass of } \text{CaO} = 2 , \text{mol} \times 56 , \text{g/mol} = 112 , \text{g} \]
Therefore, when 88 grams of carbon dioxide are released from the decomposition of 200 grams of limestone, 112 grams of calcium oxide are produced.
The correct answer is D. 112 grams.
2 answers