Question

b) in an experiment dry chlorine gas was reacted with aluminum as shown in the diagram below.

i) name substance a.
(1mark)
ii) write an equation for the reaction that took place in the combustion tube.
iii) 0.84g of aluminium reacted completely with chlorine gas. calculate the volume of chlorin gas used. (molar gas volume is 24dm³ al = 27).
(3marks)
iv) give two reasons why calcium oxide is used in the set - up.
(2marks)

Explanation:

Step1: Calculate the moles of aluminium

The molar mass of \(Al\) is \(27g/mol\). The number of moles of \(Al\), \(n(Al)=\frac{m}{M}\), where \(m = 0.84g\) and \(M = 27g/mol\). So \(n(Al)=\frac{0.84}{27}=\frac{84}{2700}=\frac{14}{450}mol\)

Step2: Use the stoichiometry of the reaction

The balanced chemical equation is \(2Al + 3Cl_{2}\xlongequal{\Delta}2AlCl_{3}\). From the equation, the mole ratio of \(Al\) to \(Cl_{2}\) is \(2:3\). Let \(n(Cl_{2})\) be the moles of \(Cl_{2}\). Then \(n(Cl_{2})=\frac{3}{2}n(Al)\)

Substitute \(n(Al)=\frac{14}{450}mol\) into the above formula: \(n(Cl_{2})=\frac{3}{2}\times\frac{14}{450}=\frac{42}{900}=\frac{7}{150}mol\)

Step3: Calculate the volume of chlorine gas

We know that \(V = n\times V_{m}\), where \(V_{m}=24dm^{3}/mol\) and \(n = n(Cl_{2})=\frac{7}{150}mol\)

So \(V=\frac{7}{150}\times24=\frac{7\times24}{150}=\frac{168}{150}= 1.12dm^{3}\)

Answer:

i) Aluminium chloride
ii) \(2Al + 3Cl_{2}\xlongequal{\Delta}2AlCl_{3}\)
iii) \(1.12dm^{3}\)
iv) - To dry the chlorine gas (remove moisture). - To absorb any un - reacted chlorine gas (prevent it from escaping to the environment as it is a toxic gas)