You need 0.0678 mol of #"Cl"_2#.
Given: Mass of #"KI"# and the chemical equation.
Find: Mass of #"Cl"_2#.
Strategy:
The central part of any stoichiometry problem is to convert moles of something to moles of something else.
(a) We start with the balanced chemical equation for the reaction.
(b) We can use the molar mass of #"KI"# to find the moles of #"KI"#.
(c) We can use the molar ratio from the equation to convert moles of #"KI"# to moles of #"Cl"_2#.
#"moles of KI" stackrelcolor(blue)("molar ratio"color(white)(Xl))(→) "moles of Cl"_2#
Our complete strategy is:
#"Mass of KI"stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of KI"stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of Cl"_2#
Solution
(a) The balanced equation is
#"Cl"_2 + "2KI" → "2KCl" + "I"_2#
(b) Calculate moles of #"KI"#
#22.5 color(red)(cancel(color(black)("g KI"))) × ("1 mol KI")/(166.00 color(red)(cancel(color(black)("g KI")))) = "0.1355 mol NH"_3#
(c) Calculate moles of #"Cl"_2#
The molar ratio of #"Cl"_2# to #"KI"# is #("1 mol Cl"_2)/("2 mol KI")"#.
#"Moles of Cl"_2 = 0.1355color(red)(cancel(color(black)("mol KI"))) × ("1 mol Cl")/(2 color(red)(cancel(color(black)("mol KI")))) = "0.0678 mol Cl"_2#
