Question

Antimony pentachloride decomposes according to this equation: SbCl5(g)⇌SbCl3(g)+Cl2(g) An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2. How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature?

Answer 1

Answer:

The mass of SbCl₅ in 2.00-L flask = 14.95 g

The mass of SbCl₃ in 2.00-L flask = 11.41 g

The mass of Cl₂ in 2.00-L flask = 3.55 g

Explanation:

Equation for the reaction is given as:

SbCl5(g)⇌SbCl3(g)+Cl2(g)

Equilibrium constant $K__C$ = $\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

Let's calculate their respective number of moles; followed by their molar concentrations.

Since number of moles $(n_x)$ = $\frac{mass}{molarmass}$

Therefore;

For SbCl₅(g)

mass given =3.85 g

molar mass = 299 g/mol

number of moles of SbCl₅(g) will be:

$n__{(SbCL_5)}$ = $\frac{3.85}{299}$

$n__{(SbCL_5)}$ = 0.0128 moles

For SbCl₃(g);

mass given = 9.14 g

molar mass = 228.13 g/mol

number of moles of SbCl₃(g) will be:

$n__{(SbCL_3)}$ = $\frac{9.14g}{228.13}$

$n__{(SbCL_3)}$ = 0.0400 moles

For Cl₂(g)

mass given  2.84 g

molar mass = 70.906 g/mol

number of moles of Cl₂(g) will be:

$n__{(Cl_2)$ = $\frac{2.84}{70.906}$

$n__{(Cl_2)$ = 0.0400 moles

Molar Concentration = $\frac{number of mole}{volume}$

Their respective molar concentration can be calculated as follows:

Molar Concentration for SbCl₅(g)

[SbCl₅(g)] = $\frac{0.0128}{5.00}$

= 0.00256

= 2.56 × 10⁻³ M

Molar Concentration for SbCl₃(g)

[SbCl₃(g)] = $\frac{0.0400}{5.00}$

= 0.008

= 8 × 10⁻³ M

Molar Concentration for Cl₂(g)

[Cl₂(g)] = $\frac{0.0400}{5.00}$

= 0.008

= 8 × 10⁻³ M

Again, we knew our Equilibrium constant $K__C$ = $\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

∴

$K__C$ = $\frac{[8*10^{-3}][8*10^{-3}]}{[2.56*10^{-3}]}$

$K__C$ = 0.025

For the equilibrium mixture in 5.00 L  flask at 448°C; $K__C$ = 0.025

∴ For 2.00 L;

Let (y) M be the concentration of SbCl₅(g), SbCl₃(g) and Cl₂(g)

∴

$K__C$ = $\frac{[y][y]}{[y]}$

$K__C$ = y

0.0025 = y

∴ For SbCl₅(g), SbCl₃(g) and Cl₂(g) in 2.00-L; their respective Molar Concentration is 0.025 M each.

So, since numbers of moles = $\frac{mass}{molarmass}$

mass = numbers of moles × molar mass

Also, numbers of moles = Molar Concentration × Volume

∴ mass = Molar Concentration × Volume × molar mass

For SbCl₅(g);

mass =  0.025 × 2 × 229g

= 14.95 g

For SbCl₃(g);

mass =  0.025 × 2 × 228.13

= 11.4065 g

≅ 11.41 g

For Cl₂(g):

mass = 0.025 × 2 × 70.906

= 3.5453 g

≅ 3.55 g

∴

The mass of SbCl₅ in 2.00-L flask = 14.95 g

The mass of SbCl₃ in 2.00-L flask = 11.41 g

The mass of Cl₂ in 2.00-L flask = 3.55 g