Question

If \(30.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{CaCl}_{2}\) is added to \(15.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3},\) what is the mass in grams of \(\mathrm{AgCl}\) precipitate?

Short answer

0.215 g of AgCl is formed.

Step-by-step solution

Determine moles of CaCl2

First, calculate the moles of \(\mathrm{CaCl}_2\). Given the volume is \(30.0\, \mathrm{mL}\) and the concentration is \(0.150\, \mathrm{M}\), use the formula \(\text{moles} = \text{concentration} \times \text{volume in liters}\). Convert \(30.0\, \mathrm{mL}\) to liters by dividing by 1000. Therefore: \[\text{moles of } \mathrm{CaCl}_2 = 0.150 \times \frac{30.0}{1000} = 0.0045\, \text{moles}\]

Determine moles of AgNO3

Calculate the moles of \(\mathrm{AgNO}_3\). With a volume of \(15.0\, \mathrm{mL}\) and a concentration of \(0.100\, \mathrm{M}\), convert the volume to liters as before: \[\text{moles of } \mathrm{AgNO}_3 = 0.100 \times \frac{15.0}{1000} = 0.0015\, \text{moles}\]

Write the balanced chemical equation

The balanced chemical reaction is: \[\mathrm{CaCl}_2 + 2\mathrm{AgNO}_3 \rightarrow 2\mathrm{AgCl} + \mathrm{Ca(NO}_3)_2\] This equation shows that 1 mole of \(\mathrm{CaCl}_2\) reacts with 2 moles of \(\mathrm{AgNO}_3\) to form 2 moles of \(\mathrm{AgCl}\).

Determine limiting reactant

Using the stoichiometry from the balanced equation, 0.0045 moles of \(\mathrm{CaCl}_2\) would require \(0.0045 \times 2 = 0.009\) moles of \(\mathrm{AgNO}_3\) to completely react. Since only 0.0015 moles of \(\mathrm{AgNO}_3\) are available, \(\mathrm{AgNO}_3\) is the limiting reactant.

Calculate moles of AgCl formed

Since \(\mathrm{AgNO}_3\) is the limiting reactant, and it reacts in a 1:1 mole ratio to form \(\mathrm{AgCl}\), the moles of \(\mathrm{AgCl}\) formed will be equal to the initial moles of \(\mathrm{AgNO}_3\): \[\text{moles of } \mathrm{AgCl} = 0.0015\, \text{moles}\]

Calculate mass of AgCl

To find the mass of \(\mathrm{AgCl}\) formed, use the formula \(\text{mass} = \text{moles} \times \text{molar mass}\). The molar mass of \(\mathrm{AgCl}\) is approximately 143.32 g/mol. Therefore: \[\text{mass of } \mathrm{AgCl} = 0.0015 \times 143.32 = 0.21498\, \text{g}\]

Key concepts

Limiting Reactant

In chemistry, determining the limiting reactant is crucial in calculating the quantities of products formed in a reaction. When two or more reactants are involved in a chemical reaction, the reactant that runs out first limits the amount of product that can be formed. This is known as the limiting reactant. For the given problem, we analyzed the moles of \(\mathrm{CaCl}_2\)and\(\mathrm{AgNO}_3\)used in the precipitation reaction. The balanced equation shows that 1 mole of\(\mathrm{CaCl}_2\)requires 2 moles of\(\mathrm{AgNO}_3\)to react completely. Here, while there are enough moles of\(\mathrm{CaCl}_2\), only 0.0015 moles of\(\mathrm{AgNO}_3\)are available, thus making\(\mathrm{AgNO}_3\)the limiting reactant. Knowing the limiting reactant helps in accurately predicting how much product is formed.

Chemical Reactions

A chemical reaction involves the transformation of reactants into products. This transformation is represented by a balanced chemical equation, which shows the relationship between the reactants and products. In the problem given, the reaction is between\(\mathrm{CaCl}_2\)and\(\mathrm{AgNO}_3\). The chemical equation is\(\mathrm{CaCl}_2 + 2\mathrm{AgNO}_3 \rightarrow 2\mathrm{AgCl} + \mathrm{Ca(NO}_3)_2\). This equation tells us that one mole of\(\mathrm{CaCl}_2\)reacts with two moles of\(\mathrm{AgNO}_3\)to produce two moles of\(\mathrm{AgCl}\)and one mole of\(\mathrm{Ca(NO}_3)_2\). Chemical reactions are fundamental in understanding how substances combine and change to form new substances. Accurately writing and balancing equations is critical for studying stoichiometry.

Precipitation Reaction

Among the different types of chemical reactions, precipitation reactions involve the formation of a solid from a solution. These reactions occur when ions in a solution combine to form a new, insoluble compound that precipitates out of the solution. In our problem, the reaction between\(\mathrm{CaCl}_2\)and\(\mathrm{AgNO}_3\)leads to the formation of solid silver chloride,\(\mathrm{AgCl}\), which is an example of a precipitate. Precipitation reactions are vital in various fields, especially in chemistry, for separating components in a mixture and for applications in qualitative analysis. Recognizing a precipitation reaction involves knowing the solubility rules and predicting which ionic compounds will form a solid.

Molar Mass Calculation

Calculating the molar mass of compounds accurately is essential for stoichiometry. The molar mass refers to the mass of one mole of a substance, expressed in grams per mole. It provides a bridge to relate moles to grams, which is necessary for determining how much reactant is required or how much product is formed. For example, in our exercise, we found the mass of\(\mathrm{AgCl}\)formed using its molar mass. The molar mass of\(\mathrm{AgCl}\)is about 143.32 g/mol. Thus, knowing the moles of\(\mathrm{AgCl}\)(0.0015 moles), we calculated:\[\text{mass of } \mathrm{AgCl} = 0.0015 \times 143.32 = 0.21498 \, \text{grams}\]Molar mass calculations form a critical part of stoichiometric calculations, guiding how mass and mole quantities are balanced in chemical equations.