The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.
Further Explanation:
To solve this problem, follow the steps below:
- Write the balanced chemical equation for the given reaction.
- Convert the mass of calcium carbonate into moles.
- Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
- Convert the number of moles of calcium oxide into mass.
Solving the given problem using the steps above:
STEP 1: The balanced chemical equation for the given reaction is:
[tex]CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}[/tex]
STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.
[tex]mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}[/tex]
STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.
For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,
[tex]mol \ CaO \ = 0.2498 \ mol[/tex]
STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.
[tex]mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g[/tex]
Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.
Therefore,
[tex]\boxed {mass \ CaO \ = 14 \ g}[/tex]
Learn More
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Keywords: thermal decomposition, stoichiometry
