The interaction you're describing can be represented by the following chemical equation:
\[ \text{CaCO}_3 , (s) \rightarrow \text{CaO} , (s) + \text{CO}_2 , (g) \]
In this reaction:
- 112 grams of calcium oxide (CaO) is produced.
- 88 grams of carbon dioxide (CO₂) is produced.
To determine the amount of calcium carbonate (CaCO₃) used in this interaction, we need to consider the molar masses of the substances involved.
The molar masses are approximately:
- CaCO₃: 100 g/mol (Calcium = 40 g/mol, Carbon = 12 g/mol, Oxygen = 16 g/mol × 3)
- CaO: 56 g/mol (Calcium = 40 g/mol, Oxygen = 16 g/mol)
- CO₂: 44 g/mol (Carbon = 12 g/mol, Oxygen = 16 g/mol × 2)
From the balanced equation, we can see that 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂.
Now, we can find the number of moles of CaO and CO₂ produced:
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Moles of CaO produced: \[ \text{Moles of CaO} = \frac{112 , \text{g}}{56 , \text{g/mol}} = 2 , \text{mol} \]
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Moles of CO₂ produced: \[ \text{Moles of CO₂} = \frac{88 , \text{g}}{44 , \text{g/mol}} = 2 , \text{mol} \]
Since the reaction shows a 1:1:1 molar ratio, that means 2 moles of CaCO₃ were used to produce 2 moles of CaO and 2 moles of CO₂.
Now we can calculate the mass of CaCO₃ used: \[ \text{Mass of CaCO₃} = \text{Moles of CaCO₃} \times \text{Molar mass of CaCO₃} = 2 , \text{mol} \times 100 , \text{g/mol} = 200 , \text{g} \]
Thus, the correct estimate of the amount of calcium carbonate used in the interaction is: Sum of 112 g and 88 g (200 g).
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