When solutions of silver nitrate and potassium chloride are mixed, silver chloride precipitates out of solution according to the equation AgNO3(aq)+KCl(aq)→AgCl(s) KNO3(aq) Part A What mass of silver chloride can be produced from 1.88 L of a 0.160 M solution of silver nitrate? Express your answer with the appropriate units
The Correct Answer and Explanation is :
To determine the mass of silver chloride (AgCl) produced when silver nitrate (AgNO₃) reacts with potassium chloride (KCl), we need to go through several steps involving stoichiometry, starting from the balanced chemical equation:
[
\text{AgNO₃ (aq)} + \text{KCl (aq)} \rightarrow \text{AgCl (s)} + \text{KNO₃ (aq)}
]
This reaction shows a 1:1 molar ratio between silver nitrate and silver chloride. Thus, one mole of silver nitrate produces one mole of silver chloride.
Step 1: Calculate the moles of AgNO₃
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. We are given:
- Volume of AgNO₃ solution = 1.88 L
- Molarity of AgNO₃ solution = 0.160 M
The number of moles of AgNO₃ can be calculated using the formula:
[
\text{moles of AgNO₃} = M \times V
]
[
\text{moles of AgNO₃} = 0.160 \, \text{mol/L} \times 1.88 \, \text{L} = 0.3008 \, \text{mol}
]
Step 2: Determine the moles of AgCl produced
From the balanced equation, we know that for every mole of AgNO₃, one mole of AgCl is produced. Therefore, the moles of AgCl produced will also be 0.3008 mol.
Step 3: Calculate the mass of AgCl
Now, to find the mass of silver chloride, we use the molar mass of AgCl. The molar mass of AgCl is calculated as:
- Molar mass of Ag (silver) = 107.87 g/mol
- Molar mass of Cl (chlorine) = 35.45 g/mol
Thus, the molar mass of AgCl is:
[
\text{Molar mass of AgCl} = 107.87 + 35.45 = 143.32 \, \text{g/mol}
]
Now, using the formula:
[
\text{mass of AgCl} = \text{moles of AgCl} \times \text{molar mass of AgCl}
]
[
\text{mass of AgCl} = 0.3008 \, \text{mol} \times 143.32 \, \text{g/mol} = 43.12 \, \text{g}
]
Final Answer:
The mass of silver chloride (AgCl) that can be produced from 1.88 L of a 0.160 M solution of silver nitrate is 43.12 grams.
Explanation:
In this process, we used stoichiometry to convert the volume and concentration of the silver nitrate solution into moles. The mole-to-mole ratio from the balanced equation tells us that the moles of AgNO₃ will be equivalent to the moles of AgCl produced. By multiplying the moles of AgCl by its molar mass, we determined the mass of the silver chloride that precipitates from the solution.