# Question 34e28

Apr 8, 2017

You need 0.0678 mol of ${\text{Cl}}_{2}$.

Given: Mass of $\text{KI}$ and the chemical equation.

Find: Mass of ${\text{Cl}}_{2}$.

Strategy:

The central part of any stoichiometry problem is to convert moles of something to moles of something else.

(b) We can use the molar mass of $\text{KI}$ to find the moles of $\text{KI}$.

(c) We can use the molar ratio from the equation to convert moles of $\text{KI}$ to moles of ${\text{Cl}}_{2}$.

${\text{moles of KI" stackrelcolor(blue)("molar ratio"color(white)(Xl))(→) "moles of Cl}}_{2}$

Our complete strategy is:

${\text{Mass of KI"stackrelcolor (blue)("molar mass"color(white)(Xl))(→) "moles of KI"stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of Cl}}_{2}$

Solution

(a) The balanced equation is

${\text{Cl"_2 + "2KI" → "2KCl" + "I}}_{2}$

(b) Calculate moles of $\text{KI}$

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 ${\text{Cl}}_{2}$

The molar ratio of ${\text{Cl}}_{2}$ to $\text{KI}$ is ("1 mol Cl"_2)/("2 mol KI")"#.

${\text{Moles of Cl"_2 = 0.1355color(red)(cancel(color(black)("mol KI"))) × ("1 mol Cl")/(2 color(red)(cancel(color(black)("mol KI")))) = "0.0678 mol Cl}}_{2}$