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Answer :
Final answer:
A metric ton of limestone containing 94.6% CaCO3 can produce 946 kg of quicklime (CaO) when heated.
Explanation:
To calculate the amount of quicklime that can be prepared by heating a metric ton of limestone, we need to consider the molar mass of calcium carbonate (CaCO3) and calcium oxide (CaO). The given limestone contains 94.6% CaCO3. We can calculate the amount of CaCO3 in a metric ton of limestone by using the percentage purity and the molar mass of CaCO3. Then, we can use stoichiometry to determine the amount of quicklime (CaO) produced from that amount of CaCO3.
Step 1: Calculate the amount of CaCO3 in a metric ton of limestone.
Amount of CaCO3 = (Purity of CaCO3 / 100) x Mass of limestone
= (94.6 / 100) x 1000 kg
= 946 kg
Step 2: Calculate the amount of CaO produced from the amount of CaCO3.
From the balanced chemical equation: CaCO3 -> CaO + CO2
We know that the molar ratio between CaCO3 and CaO is 1:1.
Therefore, the amount of CaO produced = Amount of CaCO3 = 946 kg
So, a metric ton of limestone containing 94.6% CaCO3 can produce 946 kg of quicklime (CaO).
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Final answer:
To calculate the amount of quicklime that can be prepared by heating a metric ton of limestone containing 94.6% CaCO3, multiply the weight of the limestone by the percentage of CaCO3. The amount of quicklime that can be prepared is also 0.946 metric tons.
Explanation:
To calculate the amount of quicklime that can be prepared by heating a metric ton of limestone containing 94.6% CaCO3, we need to first determine the amount of CaCO3 present in the limestone.
Since the limestone is almost pure calcium carbonate, we can assume that 94.6% of the metric ton is CaCO3.
To calculate the amount of CaCO3, we multiply the weight of the limestone (1 metric ton) by the percentage of CaCO3 (94.6%):
Amount of CaCO3 = 1 metric ton x 94.6% = 0.946 metric tons
Therefore, the amount of quicklime that can be prepared is also 0.946 metric tons, since calcium carbonate reacts completely to produce calcium oxide (quicklime) and carbon dioxide.