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?

1 Answer
Jun 13, 2017

Answer:

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.