Certainly! Let's go through the solution step by step.
1. Determine the volume and concentration of the silver nitrate ([tex]\( \text{AgNO}_3 \)[/tex]) solution:
- Volume = 1.61 liters
- Concentration = 0.158 M (molarity, which is moles per liter)
2. Calculate the number of moles of [tex]\( \text{AgNO}_3 \)[/tex] in the solution:
Since molarity (M) is the number of moles of solute per liter of solution, we can find the moles of [tex]\( \text{AgNO}_3 \)[/tex] using the formula:
[tex]\[
\text{Moles of } \text{AgNO}_3 = \text{Concentration (M)} \times \text{Volume (L)} = 0.158 \, \text{M} \times 1.61 \, \text{L} = 0.25438 \, \text{moles}
\][/tex]
3. Determine the stoichiometry of the reaction:
The balanced chemical reaction between silver nitrate ([tex]\( \text{AgNO}_3 \)[/tex]) and sodium chloride ([tex]\( \text{NaCl} \)[/tex]) is:
[tex]\[
\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3
\][/tex]
According to the reaction, 1 mole of [tex]\( \text{AgNO}_3 \)[/tex] reacts with 1 mole of [tex]\( \text{NaCl} \)[/tex] to produce 1 mole of [tex]\( \text{AgCl} \)[/tex].
4. Calculate the moles of [tex]\( \text{AgCl} \)[/tex] produced:
Since the reaction produces one mole of [tex]\( \text{AgCl} \)[/tex] for each mole of [tex]\( \text{AgNO}_3 \)[/tex], the moles of AgNO₃ will be equal to the moles of AgCl produced:
[tex]\[
\text{Moles of } \text{AgCl} = \text{Moles of } \text{AgNO}_3 = 0.25438 \, \text{moles}
\][/tex]
5. Calculate the mass of [tex]\( \text{AgCl} \)[/tex] produced:
Using the molar mass of [tex]\( \text{AgCl} \)[/tex] (143.32 g/mol), we can find the mass:
[tex]\[
\text{Mass of } \text{AgCl} = \text{Moles of } \text{AgCl} \times \text{Molar mass of } \text{AgCl} = 0.25438 \, \text{moles} \times 143.32 \, \text{g/mol} = 36.4577416 \, \text{grams}
\][/tex]
Therefore, the mass of silver chloride (AgCl) that can be produced from 1.61 liters of a 0.158 M solution of silver nitrate is 36.46 grams (when rounded to two decimal places).
