Questions: Solution stoichiometry allows chemists to determine the volume of reactants or products involved in a chemical reaction when the chemical substances are dissolved in water. Instead of using mass to calculate the resulting amount of reactants or products, molarity and volume will be used to calculate the moles of substances. Select the missing conversion factor for the following set of calculations: Assume 0.18 L of a 1.8 M solution of potassium chloride, KCl, reacts with a 1.25 M solution of lead(II) nitrate, Pb(NO3)2, to produce lead(II) chloride, PbCl2, and potassium nitrate, KNO3. The problem requires that you determine the volume of lead(II) nitrate, Pb(NO3)2, needed for the reaction to occur. 2 KCl(aq) + Pb(NO3)2(aq) -> PbCl2(s) + 2 KNO3(aq) 1 mol KCl / 2 mol Pb(NO3)2 2 mol KCl / 1 mol Pb(NO3)2 1 mol Pb(NO3)2 / 2 mol KNO3

Solution stoichiometry allows chemists to determine the volume of reactants or products involved in a chemical reaction when the chemical substances are dissolved in water. Instead of using mass to calculate the resulting amount of reactants or products, molarity and volume will be used to calculate the moles of substances. Select the missing conversion factor for the following set of calculations: Assume 0.18 L of a 1.8 M solution of potassium chloride, KCl, reacts with a 1.25 M solution of lead(II) nitrate, Pb(NO3)2, to produce lead(II) chloride, PbCl2, and potassium nitrate, KNO3. The problem requires that you determine the volume of lead(II) nitrate, Pb(NO3)2, needed for the reaction to occur. 2 KCl(aq) + Pb(NO3)2(aq) -> PbCl2(s) + 2 KNO3(aq) 1 mol KCl / 2 mol Pb(NO3)2 2 mol KCl / 1 mol Pb(NO3)2 1 mol Pb(NO3)2 / 2 mol KNO3
Transcript text: Solution stoichiometry allows chemists to determine the volume of reactants or products involved in a chemical reaction when the chemical substances are dissolved in water. Instead of using mass to calculate the resulting amount of reactants or products, molarity and volume will be used to calculate the moles of substances. Select the missing conversion factor for the following set of calculations: Assume 0.18 L of a 1.8 M solution of potassium chloride, KCl , reacts with a 1.25 M solution of lead(II) nitrate, $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$, to produce lead(II) chloride, $\mathrm{PbCl}_{2}$, and potassium nitrate, $\mathrm{KNO}_{3}$. The problem requires that you determine the volume of lead(II) nitrate, $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$, needed for the reaction to occur. \[ 2 \mathrm{KCl}(\mathrm{aq})+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\dot{\mathrm{~s}})+2 \mathrm{KNO}_{3}(\mathrm{aq}) \] \[ \frac{1 \mathrm{~mol} \mathrm{KCl}}{2 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}} \] \[ \frac{2 \mathrm{~mol} \mathrm{KCl}}{1 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}} \] \[ \frac{1 \mathrm{~mol} \mathrm{~Pb}^{\mathrm{ma}}\left(\mathrm{NO}_{3}\right)_{2}}{2 \mathrm{~mol} \mathrm{KNO}_{3}} \]
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Solution

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Solution Steps

Step 1: Determine the moles of KCl

First, we need to calculate the moles of potassium chloride (KCl) using its volume and molarity.

\[ \text{Moles of KCl} = \text{Molarity of KCl} \times \text{Volume of KCl} \]

Given:

  • Volume of KCl = 0.18 L
  • Molarity of KCl = 1.8 M

\[ \text{Moles of KCl} = 1.8 \, \text{M} \times 0.18 \, \text{L} = 0.324 \, \text{moles} \]

Step 2: Use the stoichiometric ratio to find moles of Pb(NO\(_3\))\(_2\)

From the balanced chemical equation:

\[ 2 \mathrm{KCl}(\mathrm{aq}) + \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) + 2 \mathrm{KNO}_{3}(\mathrm{aq}) \]

The stoichiometric ratio between KCl and Pb(NO\(_3\))\(_2\) is 2:1. Therefore, the moles of Pb(NO\(_3\))\(_2\) needed can be calculated as:

\[ \text{Moles of Pb(NO}_3\text{)}_2 = \frac{0.324 \, \text{moles KCl}}{2} = 0.162 \, \text{moles} \]

Step 3: Calculate the volume of Pb(NO\(_3\))\(_2\) solution needed

Finally, we use the molarity of the Pb(NO\(_3\))\(_2\) solution to find the required volume.

\[ \text{Volume of Pb(NO}_3\text{)}_2 = \frac{\text{Moles of Pb(NO}_3\text{)}_2}{\text{Molarity of Pb(NO}_3\text{)}_2} \]

Given:

  • Molarity of Pb(NO\(_3\))\(_2\) = 1.25 M

\[ \text{Volume of Pb(NO}_3\text{)}_2 = \frac{0.162 \, \text{moles}}{1.25 \, \text{M}} = 0.1296 \, \text{L} \]

Final Answer

The missing conversion factor is:

\[ \boxed{\frac{2 \, \text{mol KCl}}{1 \, \text{mol Pb(NO}_3\text{)}_2}} \]

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