# What are the smallest and largest volumes of 0.15 mol/L silver nitrate solution required to react with 0.3 g of a mixture of potassium chloride and barium chloride dihydrate?

Jun 13, 2017

WARNING! Long answer! The smallest and largest volumes of ${\text{AgNO}}_{3}$ are 16 mL and 34 mL, respectively.

#### Explanation:

I would start by calculating the $\text{% Cl}$ in each ingredient.

color(blue)(bar(ul(|color(white)(a/a)"% Cl" = "mass of Cl"/"mass of compound"× 100 %color(white)(a/a)|)))" "

In $\text{BaCl"_2·"2H"_2"O}$,

"% Cl" = (70.91 color(red)(cancel(color(black)("u"))))/(244.26 color(red)(cancel(color(black)("u")))) × 100 % = 29.03 %

In $\text{KCl}$,

"% Cl" = (35.45 color(red)(cancel(color(black)("u"))))/(74.55 color(red)(cancel(color(black)("u")))) × 100 % = 47.55 %

In $\text{NaCl}$,

"% Cl" = (35.45 color(red)(cancel(color(black)("u"))))/(58.44 color(red)(cancel(color(black)("u")))) × 100 % = 60.66 %

Thus, $\text{NaCl}$ contains the most $\text{Cl}$ and $\text{BaCl"_2·"2H"_2"O}$ contains the least $\text{Cl}$ on the basis of mass percent.

The largest volume of $\text{AgNO"_3}$

This comes from pure $\text{NaCl}$.

The equation for the reaction is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 58.44$
$\textcolor{w h i t e}{m m m} {\text{NaCl" + "AgNO"_3 → "AgCl + NaNO}}_{3}$

${\text{Moles of AgNO}}_{3}$

= 0.3 color(red)(cancel(color(black)("g NaCl" )))× (1 color(red)(cancel(color(black)("mol NaCl"))))/(58.44 color(red)(cancel(color(black)("g NaCl")))) × "1 mol AgNO"_3/(1 color(red)(cancel(color(black)("mol NaCl")))) = "0.0051 mol AgNO"_3

${\text{Volume of AgNO"_3 = 0.0051 color(red)(cancel(color(black)("mol AgNO"_3))) × "1 L AgNO"_3/(0.15 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.034 L AgNO"_3 = "34 mL AgNO}}_{3}$

The smallest volume of $\text{AgNO"_3}$

This comes from pure $\text{BaCl"_2·2"H"_2"O}$.

The equation for the reaction is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m l l} 244.26$
$\textcolor{w h i t e}{m m m} \text{BaCl"_2·2"H"_2"O" + "2AgNO"_3 → "2AgCl + Ba(NO"_3)_2 + 2"H"_2"O}$

${\text{Moles of AgNO}}_{3}$

= 0.3 color(red)(cancel(color(black)("g BaCl"_2·2"H"_2"O" )))× (1 color(red)(cancel(color(black)("mol BaCl"_2·2"H"_2"O"))))/(244.26 color(red)(cancel(color(black)("g BaCl"_2·2"H"_2"O")))) × "2 mol AgNO"_3/(1 color(red)(cancel(color(black)("mol BaCl"_2·2"H"_2"O")))) = "0.0025 mol AgNO"_3

${\text{Volume of AgNO"_3 = 0.0025 color(red)(cancel(color(black)("mol AgNO"_3))) × "1 L AgNO"_3/(0.15 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.016 L AgNO"_3 = "16 mL AgNO}}_{3}$

Note: The answer can have only one significant figure, because that is all you gave for the mass of the sample, but I calculated the volumes to two significant figures.