Question

to solve stoichiometry problems, you must always calculate numbers of moles. recall that molarity, (m), is equal to the concentration in moles per liter (m=\text{mol}/\text{l}). what volume of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation (2\text{agno}_3(\text{aq})+\text{cacl}_2(\text{aq})
ightarrow2\text{agcl}(\text{s})+\text{ca}(\text{no}_3)_2(\text{aq})). part a: what mass of silver chloride can be produced from 1.00 l of a 0.195 m solution of silver nitrate? express your answer with the appropriate units. view available hint(s) mass of (\text{agcl}=29.0\text{ g}) previous answers correct. part b: the reaction described in part a required 3.46 l of calcium chloride. what is the concentration of this calcium chloride solution? express your answer with the appropriate units. view available hint(s)

Explanation:

Step1: Calculate moles of silver nitrate

The molarity formula is $M=\frac{n}{V}$, where $M$ is molarity, $n$ is moles and $V$ is volume in liters. Given $M = 0.195\ M$ and $V=1.00\ L$ for silver nitrate ($AgNO_3$), we can find $n$. Rearranging the formula gives $n = M\times V$. So, $n_{AgNO_3}=0.195\ mol/L\times1.00\ L = 0.195\ mol$.

Step2: Use stoichiometry

From the balanced chemical equation $2AgNO_3(aq)+CaCl_2(aq)\to2AgCl(s)+Ca(NO_3)_2(aq)$, the mole - ratio of $AgNO_3$ to $AgCl$ is $2:2$ or $1:1$. So, $n_{AgCl}=n_{AgNO_3}=0.195\ mol$.

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 have $m_{AgCl}=0.195\ mol\times143.32\ g/mol\approx28.0\ g$ (for Part A).

For Part B:

Step1: Use stoichiometry to find moles of $CaCl_2$

Since the mole - ratio of $AgNO_3$ to $CaCl_2$ is $2:1$, and $n_{AgNO_3}=0.195\ mol$, then $n_{CaCl_2}=\frac{0.195\ mol}{2}=0.0975\ mol$.

Step2: Calculate molarity of $CaCl_2$

The molarity formula is $M=\frac{n}{V}$. Given $n = 0.0975\ mol$ and $V = 3.46\ L$, then $M_{CaCl_2}=\frac{0.0975\ mol}{3.46\ L}\approx0.0282\ mol/L$.

Answer:

Part A: mass of $AgCl = 28.0\ g$
Part B: Value = $0.0282$, Units = $mol/L$