Question #5a225

1 Answer
Jul 6, 2017

Answer:

#4.13# #"g excess KCl"#

Explanation:

We know that #4.6# #"g PbCl"_2# were recovered, and that #"KCl"# is the excess reactant (according to the image).

We can convert this value to moles of #"PbCl"_2# using its molar mass (#278.1# #"g/mol"#):

#4.6cancel("g PbCl"_2)((1color(white)(l)"mol PbCl"_2)/(278.1cancel("g PbCl"_2))) = color(red)(0.0165# #color(red)("mol PbCl"_2#

We can then use the coefficients of the equation to determine the moles of #"KCl"# used up:

#color(red)(0.0165)cancel(color(red)("mol PbCl"_2))((2color(white)(l)"mol KCl")/(1cancel("mol PbCl"_2))) = color(green)(0.0331# #color(green)("mol KCl"#

Lastly (using molar mass of #"KCl"#), let's convert this value to grams, and subtract from the original value to determine how much #"KCl"# was not used:

#color(green)(0.0331)cancel(color(green)("mol KCl"))((74.55color(white)(l)"g KCl")/(1cancel("mol KCl"))) = color(purple)(2.47# #color(purple)("g KCl"#

Subtracting from original value, we have

#6.60# #"g KCl"# #- color(purple)(2.47# #color(purple)("g KCl"# #= color(blue)(4.13# #color(blue)("g excess KCl"#

which, rounded to one decimal place like you have been doing, is #color(blue)(4.1# #color(blue)("g KCl"#.