Silver nitrate can react with potassium chloride in aqueous solution to produce silver chloride, as shown: AgNO3 (aq) + KCl(aq) --> AgCl(s) + KNO3 (aq) The reaction of 230 mL of AgNO3 (aq) with excess aqueous potassium chloride produces 14.70 g of precipitate. The concentration of the AgNO3(aq) solution is _________ mol/L. answer to one digit after the decimal point
The Correct Answer and Explanation is:
To find the concentration of the AgNO3 solution, we will follow these steps:
Step 1: Write the balanced equation
The given reaction is:AgNO3(aq)+KCl(aq)→AgCl(s)+KNO3(aq)\text{AgNO}_3 (aq) + \text{KCl} (aq) \rightarrow \text{AgCl} (s) + \text{KNO}_3 (aq)AgNO3(aq)+KCl(aq)→AgCl(s)+KNO3(aq)
This shows a 1:1 molar ratio between AgNO3 and AgCl.
Step 2: Calculate moles of AgCl produced
We know the mass of AgCl precipitate is 14.70 g. To find the moles of AgCl, we use the molar mass of AgCl:Molar mass of AgCl=107.87 g/mol (Ag)+35.45 g/mol (Cl)=143.32 g/mol\text{Molar mass of AgCl} = 107.87 \, \text{g/mol (Ag)} + 35.45 \, \text{g/mol (Cl)} = 143.32 \, \text{g/mol}Molar mass of AgCl=107.87g/mol (Ag)+35.45g/mol (Cl)=143.32g/mol
Now calculate the moles of AgCl:Moles of AgCl=Mass of AgClMolar mass of AgCl=14.70 g143.32 g/mol=0.1025 mol\text{Moles of AgCl} = \frac{\text{Mass of AgCl}}{\text{Molar mass of AgCl}} = \frac{14.70 \, \text{g}}{143.32 \, \text{g/mol}} = 0.1025 \, \text{mol}Moles of AgCl=Molar mass of AgClMass of AgCl=143.32g/mol14.70g=0.1025mol
Step 3: Calculate moles of AgNO3
From the balanced equation, we know that 1 mole of AgNO3 produces 1 mole of AgCl. Therefore, the moles of AgNO3 are the same as the moles of AgCl:Moles of AgNO3=0.1025 mol\text{Moles of AgNO3} = 0.1025 \, \text{mol}Moles of AgNO3=0.1025mol
Step 4: Calculate the concentration of AgNO3
We are given the volume of AgNO3 solution as 230 mL (or 0.230 L). The concentration (C) of AgNO3 is given by:C=Moles of AgNO3Volume of solution in liters=0.1025 mol0.230 L=0.4457 mol/LC = \frac{\text{Moles of AgNO3}}{\text{Volume of solution in liters}} = \frac{0.1025 \, \text{mol}}{0.230 \, \text{L}} = 0.4457 \, \text{mol/L}C=Volume of solution in litersMoles of AgNO3=0.230L0.1025mol=0.4457mol/L
Final Answer:
The concentration of the AgNO3 solution is approximately 0.4 mol/L when rounded to one decimal place.
