91Ó°ÊÓ

Chemical reactions involve the transformation of substances through breaking and forming bonds, resulting in new products. In our exercise, we are dealing with the reaction where aluminum iodide (\(\mathrm{AlI}_3\)) reacts with silver nitrate (\(\mathrm{AgNO}_3\)) to form silver iodide (\(\mathrm{AgI}\)) and aluminum nitrate (\(\mathrm{Al(NO}_3)_3\)). This type of reaction is known as a double displacement reaction. Each reactant swaps components to form two new compounds.

This reaction can be better understood in terms of stoichiometry, which allows us to calculate how much product can form from given amounts of reactants. In this case, the balanced reaction equation is crucial. It tells us that 1 mole of \(\mathrm{AlI}_3\) reacts with 3 moles of \(\mathrm{AgNO}_3\) to produce 3 moles of \(\mathrm{AgI}\). This molar ratio is necessary for calculating how much \(\mathrm{AgI}\) will be formed from a known amount of \(\mathrm{AlI}_3\). By understanding chemical reactions and stoichiometry, we can ensure precise calculations of material usage and product yield in chemical processes.

Mole calculations are essential for translating between mass and the number of particles in chemistry. The mole concept connects the mass of a substance to the number of atoms or molecules it contains. How? By using its molar mass!

For \(\mathrm{AlI}_3\), the molar mass is approximately \(407.695 \, \mathrm{g/mol}\). To find out how many moles we have, we divide the sample mass by this value. In our exercise, we calculated the moles of \(\mathrm{AlI}_3\) as follows:

• Given mass of \(\mathrm{AlI}_3\): \(0.102912 \, \mathrm{g}\)
• Molar mass of \(\mathrm{AlI}_3\): \(407.695 \, \mathrm{g/mol}\)
• Conversion to moles: \(\frac{0.102912}{407.695} \approx 0.0002524 \, \text{mol}\)

Understanding mole calculations helps chemists determine how far a reaction will proceed and how much of each element in a compound reacts or is produced.

The concept of mass percent composition helps describe the proportions of each component within a substance. It expresses the mass of each element or compound as a percentage of the total mass. This percentage is crucial for quantitative analysis and quality control in chemical manufacturing.

In the given exercise, we started by using the mass percent composition of \(\mathrm{AlI}_3\) to find out how much of it is present in the \(0.512-\mathrm{g}\) sample. The sample assays \(20.1\% \mathrm{AlI}_3\), meaning simply:

• Multiply the total sample mass by the mass percent to find the mass of \(\mathrm{AlI}_3\): \(0.512 \, \mathrm{g} \times \frac{20.1}{100} \approx 0.102912 \, \mathrm{g}\).

This technique is often used in analytical chemistry to calculate how much of a particular substance is in a mixture. It gives critical insights into the composition of materials, ensuring the desired properties or purity in applications.