Question

2AlCl₃    =>   2Al    +    3Cl₂

If 533.1 grams of AlCl₃ decomposed at 37.0 °C and 3.02 atm, what would be the volume of chlorine gas?

Answer 1

Moles of AlCl₃ decomposed;

$moles = \dfrac{mass}{MM}=\dfrac{533.1g}{133.34g/mol}=3.998mol$

The balanced equation for the reaction;

2AlCl₃ $\to$ 2Al + 3Cl₂

Moles of Cl₂ produced;

Since the mole ratio of AlCl₃ to Cl₂ is 2:3, then

Moles of Cl₂ $=\dfrac{3}{2}xmoles of AlCl3$

$=\dfrac{3}{2}x3.998mol$

$=5.997mol$

Now, at STP, the molar volume of an ideal gas is 22.4L

Assuming ideal situation;

If 1mole of Cl₂ will occupy 22.4L

5.997 moles of Cl₂ will occupy?

$=\dfrac{5.997molx22.4L}{1mol} = 134.334L$

This is at STP (T₁ = 273K, P₁ = 1 atm, V₁ = 134.334L)

At the new conditions (T₂ = 273 + 37 =310K, P₂ = 3.02 atm, V₂ = ?)

$\dfrac{P1V1}{T1}=\dfrac{P2V2}{T2}$

$P2 = \dfrac{P1V1T2}{T1} = \dfrac{1atm x 134.334Kx310L}{237K}$

= 152.54L