# If "10 g" of "AlCl"_3" decomposes, how many grams of "Cl"_2" will be produced?

May 1, 2017

This is a stoichiometry problem.

#### Explanation:

First, you need to find the number of moles of aluminum chloride.

moles = mass/rmm

$\frac{10 g}{26.98 + 3 \left(35.453\right)}$ = 0.075 moles

Since the equation states that for every 3 molecules of aluminum chloride, you produce 2 molecules of chlorine, by cross-multiplying the amount of moles, you obtain 0.1125 moles of chlorine produced.

Now, just convert moles to grams by rearranging the formula earlier, and you should obtain 7.98 grams (rounded to 2 decimal places)

May 1, 2017

$\approx \text{8 g Cl"_2}$ will be produced

#### Explanation:

Balanced Equation

${\text{2AlCl}}_{3}$$\rightarrow$${\text{2Al + 3Cl}}_{2}$

The process for solving this problem is:

color(red)("given mass AlCl"_3"$\rightarrow$color(blue)("mol AlCl"_3"$\rightarrow$color(green)("mol Cl"_2"$\rightarrow$color(purple) ("mass Cl"_2"

The molar masses of aluminum chloride and chlorine gas must be determined. Molar mass is determined by multiplying the subscript of each element by its atomic weight (relative atomic mass) from the periodic table in grams/mole, or g/mol.

Molar Masses
${\text{AlCl}}_{3} :$(1xx26.982"g/mol Al")+(3xx35.45"g/mol Cl")="133.332 g/mol AlCl"_3"
(I will round to the correct number of significant figures at the end.

${\text{Cl}}_{2} :$(2xx35.45"g/mol Cl")="70.9 g/mol Cl"_2"

Mole Ratios
Write the mole ratios between $\text{AlCl"_3}$ and $\text{Cl"_2}$ from the balanced equation. The mole ratio allows us to convert between moles and mass and vice-versa.

$\left(2 {\text{mol AlCl"_3)/(3"mol Cl}}_{2}\right)$ $\textcolor{w h i t e}{\ldots .} \text{and}$ $\textcolor{w h i t e}{\ldots .} \left(3 {\text{mol Cl"_2)/(2"mol AlCl}}_{3}\right)$

color(red)("Given Mass" color(red)("AlCl"_3"$\rightarrow$color(blue)("Mol AlCl"_3"

Multiply the given mass of $\text{AlCl"_3}$ by the reciprocal of the molar mass of $\text{Cl"_2}$.

10color(red)cancel(color(black)("g AlCl"_3))xx(1"mol AlCl"_3)/(133.332color(red)cancel(color(black)("g AlCl"_3)))="0.0750008 mol AlCl"_3"

color(blue)("Mol AlCl"_3"$\rightarrow$color(green)("Mol Cl"_2"

Multiply mol $\text{AlCl"_3}$ by the mole ratio with $\text{Cl"_2}$ in the numerator.

0.0750008color(red)cancel(color(black)("mol AlCl"_3))xx(3"mol Cl"_2)/(2color(red)cancel(color(black)("mol AlCl"_3)))="0.11250 mol Cl"_2"

color(green)("Mole Cl"_2"$\rightarrow$$\textcolor{p u r p \le}{\text{Mass Cl"_2}}$

Multiply mol $\text{Cl"_2}$ by its molar mass.

0.11250color(red)cancel(color(black)("mol Cl"_2))xx(70.9"g Cl"_2)/(1color(red)cancel(color(black)("mol Cl"_2)))="7.976 g Cl"_2~~"8 g Cl"_2" (rounded to one significant figure due to 10 g $\text{AlCl"_3}$