Calculate the mass of the precipitate formed when 250. mL of a 0.100 M aqueous solution of silver nitrate reacts with 200. mL of a 0.150 M aqueous solution of sodium chloride. AgNO3(aq) + NaCI(aq) AgCI(s) + NaN03(aq)
Calculate the mass of the precipitate formed when 250. mL of a 0.100 M aqueous solution of silver nitrate reacts with 200. mL of a 0.150 M aqueous solution of sodium chloride. AgNO3(aq) + NaCI(aq) AgCI(s) + NaN03(aq)
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![### Test Bank, Question 14.120
**Problem Statement:**
Calculate the mass of the precipitate formed when 250. mL of a 0.100 M aqueous solution of silver nitrate reacts with 200. mL of a 0.150 M aqueous solution of sodium chloride.
**Chemical Equation:**
\[ \text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq) \]
**Solution:**
1. Identify the limiting reactant by calculating the moles of each reactant:
- Moles of \(\text{AgNO}_3\):
\[ \text{Molarity} \times \text{Volume (in L)} = 0.100 \, \text{M} \times 0.250 \, \text{L} = 0.025 \, \text{moles} \]
- Moles of \(\text{NaCl}\):
\[ \text{Molarity} \times \text{Volume (in L)} = 0.150 \, \text{M} \times 0.200 \, \text{L} = 0.030 \, \text{moles} \]
2. Since \(\text{AgNO}_3\) forms a 1:1 stoichiometric ratio with \(\text{NaCl}\), and \(0.025 \, \text{moles} < 0.030 \, \text{moles}\), \(\text{AgNO}_3\) is the limiting reactant.
3. Calculate the moles of \(\text{AgCl}\) formed, which is equal to the moles of the limiting reactant (\(\text{AgNO}_3\)):
\[ \text{Moles of } \text{AgCl} = 0.025 \, \text{moles} \]
4. The molar mass of \(\text{AgCl}\) (Silver chloride) is:
\[ \text{Ag: } 107.87 \, \text{g/mol} + \text{Cl: } 35.45 \, \text{g/mol} = 143.32 \, \text{g/mol} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2f9e96de-8215-4532-a7d0-6f1fdf87f241%2F9654a537-9212-40cd-ba45-ac718833ca69%2Fdwwv1pc.png&w=3840&q=75)
Transcribed Image Text:### Test Bank, Question 14.120
**Problem Statement:**
Calculate the mass of the precipitate formed when 250. mL of a 0.100 M aqueous solution of silver nitrate reacts with 200. mL of a 0.150 M aqueous solution of sodium chloride.
**Chemical Equation:**
\[ \text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq) \]
**Solution:**
1. Identify the limiting reactant by calculating the moles of each reactant:
- Moles of \(\text{AgNO}_3\):
\[ \text{Molarity} \times \text{Volume (in L)} = 0.100 \, \text{M} \times 0.250 \, \text{L} = 0.025 \, \text{moles} \]
- Moles of \(\text{NaCl}\):
\[ \text{Molarity} \times \text{Volume (in L)} = 0.150 \, \text{M} \times 0.200 \, \text{L} = 0.030 \, \text{moles} \]
2. Since \(\text{AgNO}_3\) forms a 1:1 stoichiometric ratio with \(\text{NaCl}\), and \(0.025 \, \text{moles} < 0.030 \, \text{moles}\), \(\text{AgNO}_3\) is the limiting reactant.
3. Calculate the moles of \(\text{AgCl}\) formed, which is equal to the moles of the limiting reactant (\(\text{AgNO}_3\)):
\[ \text{Moles of } \text{AgCl} = 0.025 \, \text{moles} \]
4. The molar mass of \(\text{AgCl}\) (Silver chloride) is:
\[ \text{Ag: } 107.87 \, \text{g/mol} + \text{Cl: } 35.45 \, \text{g/mol} = 143.32 \, \text{g/mol} \
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