what mass of silver chloride can be produced from 1.81 l of a 0.182 m solution of silver nitrate? express…

Explanation:

Step1: Calculate moles of silver nitrate

The formula for molarity $M=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume in liters. Rearranging for $n$, we have $n = M\times V$. Given $M = 0.182\ M$ and $V=1.81\ L$, then $n_{AgNO_3}=0.182\ mol/L\times1.81\ L = 0.32942\ mol$.

Step2: Determine moles of silver chloride

The chemical reaction between silver nitrate ($AgNO_3$) and potassium chloride ($KCl$) is $AgNO_3 + KCl=AgCl\downarrow+KNO_3$. The mole - ratio of $AgNO_3$ to $AgCl$ is 1:1. So, $n_{AgCl}=n_{AgNO_3}=0.32942\ mol$.

Step3: Calculate mass of silver chloride

The molar mass of $AgCl$ is $M_{AgCl}=107.87\ g/mol + 35.45\ g/mol=143.32\ g/mol$. Using the formula $m = n\times M$, we get $m_{AgCl}=0.32942\ mol\times143.32\ g/mol\approx47.2\ g$.

Step4: For part B, find moles of KCl

Since the mole - ratio of $AgNO_3$ to $KCl$ is 1:1 in the reaction, $n_{KCl}=n_{AgNO_3}=0.32942\ mol$.

Step5: Calculate molarity of KCl solution

Using the molarity formula $M=\frac{n}{V}$, with $n = 0.32942\ mol$ and $V = 3.02\ L$, we have $M_{KCl}=\frac{0.32942\ mol}{3.02\ L}\approx0.109\ M$.

Answer:

Part A: Value: 47.2 Units: g Part B: Value: 0.109 Units: M