Question

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?

Answer 1

WARNING! Long answer! The smallest and largest volumes of `"AgNO"_3` are 16 mL and 34 mL, respectively.

Explanation:

I would start by calculating the `"% 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 `"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 `"KCl"`,

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

In `"NaCl"`,

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

Thus, `"NaCl"` contains the most `"Cl"` and `"BaCl"_2·"2H"_2"O"` contains the least `"Cl"` on the basis of mass percent.

The largest volume of `"AgNO"_3"`

This comes from pure `"NaCl"`.

The equation for the reaction is

`M_text(r):color(white)(m)58.44`
`color(white)(mmm)"NaCl" + "AgNO"_3 → "AgCl + NaNO"_3`

`"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`

`"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 `"AgNO"_3"`

This comes from pure `"BaCl"_2·2"H"_2"O"`.

The equation for the reaction is

`M_text(r):color(white)(mmll)244.26`
`color(white)(mmm)"BaCl"_2·2"H"_2"O" + "2AgNO"_3 → "2AgCl + Ba(NO"_3)_2 + 2"H"_2"O"`

`"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`

`"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.