91Ó°ÊÓ

In any chemical reaction, the limiting reactant is the substance that is completely consumed first and therefore limits the amount of products formed. Understanding which reactant is limiting is key to calculating the quantities of products and any excess reactants left after the reaction. In this exercise, we are mixing two substances:

• Silver nitrate (\(\mathrm{AgNO}_3\)
• Potassium chromate (\(\mathrm{K}_2\mathrm{CrO}_4\)

These react to form a precipitate, \(\mathrm{Ag}_2\mathrm{CrO}_4\). To find the limiting reactant, we compare the mole ratio from the balanced chemical equation:

• 2 moles of \(\mathrm{AgNO}_3\) react with 1 mole of \(\mathrm{K}_2\mathrm{CrO}_4\).

By calculating the moles of each reactant present in the solution, we determine that \(\mathrm{AgNO}_3\) is the limiting reactant. This is because 0.00294 moles of \(\mathrm{AgNO}_3\) are available, which is less than the 0.00308 moles required to fully react with \(\mathrm{K}_2\mathrm{CrO}_4\). As a result, all available \(\mathrm{AgNO}_3\) will be used up first.

A balanced chemical equation is crucial for understanding how reactants interact to form products. It provides the relative number of moles needed for the reaction and helps ensure mass conservation. For the reaction in this problem, the balanced chemical equation is: \[2 \mathrm{AgNO}_3 + \mathrm{K}_2\mathrm{CrO}_4 \rightarrow \mathrm{Ag}_2\mathrm{CrO}_4 \downarrow + 2 \mathrm{KNO}_3\]This equation tells us that two moles of silver nitrate react with one mole of potassium chromate to produce one mole of silver chromate and two moles of potassium nitrate.

• The coefficients (numbers before substances) indicate how many moles of each substance participate in the chemical reaction.
• Balancing the equation ensures that the number of atoms for each element is the same on both sides, preserving the law of conservation of mass.

In this case, having balanced coefficients allows us to determine the exact proportions of each reactant needed and to identify the limiting reactant.

Mole calculations help quantify the amounts of reactants and products involved in a chemical reaction. The mole is a fundamental unit in chemistry, linking mass to the number of chemical entities (like atoms or molecules). To solve this problem, we start by determining the moles of each reactant:

• The molar mass of \(\mathrm{AgNO}_3\) is 169.87 g/mol. For 0.500 g of \(\mathrm{AgNO}_3\), the moles are calculated as \(\frac{0.500}{169.87} \approx 0.00294\).
• The molar mass of \(\mathrm{K}_2\mathrm{CrO}_4\) is 194.19 g/mol. For 0.300 g of \(\mathrm{K}_2\mathrm{CrO}_4\), the moles are \(\frac{0.300}{194.19} \approx 0.00154\).

Using these mole calculations, we can identify the limiting reactant and predict the mass of the precipitate formed. We also use them to find the mass of any leftover reactant. Once the limiting reactant \(\mathrm{AgNO}_3\) is fully consumed, we use the balanced equation to find the moles of \(\mathrm{Ag}_2\mathrm{CrO}_4\) formed: \(\frac{0.00294}{2} = 0.00147\). Knowing its molar mass, 331.73 g/mol, we find that about 0.488 g of \(\mathrm{Ag}_2\mathrm{CrO}_4\) is produced as precipitate.