91Ó°ÊÓ

Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist in chemical reactions. It helps us understand how much of a reactant is needed to produce a certain amount of product. In the exercise, stoichiometry is used to relate the amount of \(\text{CaCO}_3\) decomposed to the amount of \(\text{CO}_2\) produced.

To apply stoichiometry here, we started by writing the balanced chemical equation for the decomposition: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
From the balanced equation, we know:

• 1 mole of \(\text{CaCO}_3\) decomposes to form 1 mole of \(\text{CO}_2\).

We then calculated the moles of \(\text{CO}_2\) produced from the given mass using the formula:
\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]
From this, the exercise demonstrated that understanding stochiometry allowed us to determine the corresponding amount of \(\text{CaCO}_3\) present in the original sample.

Mass percent is a way of expressing the concentration of a component in a mixture. It is defined as the mass of the component divided by the total mass of the mixture, multiplied by 100 to get a percentage.

In our exercise, we needed to find out the mass percent of \(\text{CaCO}_3\) in the original mixture. We followed these steps:

• First, calculate the mass of \(\text{CaCO}_3\) from the moles using: \(\text{mass} = \text{moles} \times \text{molar mass}\)
• Then, use the formula for mass percent:
  \text{Mass percent} = \left( \frac{\text{mass of component}}{\text{total mass of mixture}} \right) \times 100\

This calculation revealed that the mass percent of \(\text{CaCO}_3\) in the mixture was 60.6%. This way, mass percent helps in identifying the concentration of \(\text{CaCO}_3\) accurately.

Chemical reactions involve the transformation of reactants into products by breaking and forming bonds. The exercise focused on a decomposition reaction where calcium carbonate (\text{CaCO}_3) breaks down into calcium oxide (\text{CaO}) and carbon dioxide (\text{CO}_2).
This reaction can be written as: \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\

In this process:

• \text{CaCO}_3 is the reactant.
• \text{CaO} and \text{CO}_2 are the products.

Understanding the nature of the reaction allowed us to determine the masses and moles of products formed. It also showed the importance of balancing chemical equations, as it provides the necessary ratios to solve the problem efficiently.
In the context of the exercise, noticing that the mass of the remaining solid after heating was less helped us figure out that \(\text{CaCO}_3\) decomposed completely into \(\text{CO}_2\) gas that escaped, as well as into \(\text{CaO}\). By recognizing these changes, we could accurately find out the original mass and, consequently, the mass percentage of the initial reactant.