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If 15.2 g of aluminum reacts with 39.1 g of chlorine, how many grams of AlCl₃ form?

\[ 2Al + 3Cl_2 \rightarrow 2AlCl_3 \]

Answer :

Answer

Mass of AlCl3 formed = 75.12 g

Explanation

Given:

2Al + 3Cl2 --> 2 AlCl3

Mass of aluminum = 15.2 g

Mass of chlorine = 39.1 g

Required: How many grams of AlCl3 forms

Solution

Aluminum

[tex]\begin{gathered} 15.2\text{ g Al x }\frac{1\text{ mole Al}}{26,98\text{ g Al}}\text{ x }\frac{2\text{ mole AlCl}_3\text{ }}{2\text{ mole Al}}\text{ x }\frac{133,34\text{ g AlCl}_3}{1\text{ mole AlCl}_3} \\ \\ =\text{ 75.12 g AlCl}_3 \end{gathered}[/tex]

Chlorine

[tex]\begin{gathered} 39.1\text{ g Cl x }\frac{1\text{ mole Cl}}{35,45\text{ g Cl}}\text{ x }\frac{2\text{ mole AlCl}_3}{3\text{ mole Cl}_2}\text{ x }\frac{133.34\text{ g AlCl}_3}{1\text{ mole AlCl}_3} \\ \\ =\text{ 98.05 g AlCl}_3 \end{gathered}[/tex]

Aluminum is the limiting reagent because it produces less mass of AlCl3. Therefore the mass of AlCl3 that will be formed = 75.12 g

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