QuestionMay 7, 2025

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al+3Cl_(2)-2AlCl_(3) How many grams of aluminum chloride could be produced by reacting 34.0 g of aluminum and 39.0 g of chlorine gas? __ g

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al+3Cl_(2)-2AlCl_(3) How many grams of aluminum chloride could be produced by reacting 34.0 g of aluminum and 39.0 g of chlorine gas? __ g
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al+3Cl_(2)-2AlCl_(3) How many grams of aluminum chloride could be produced by reacting 34.0 g of aluminum and 39.0 g of chlorine gas? __ g

Solution
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Answer

49.33 g Explanation 1. Determine molar masses Molar mass of Al = 26.98 \, g/mol, Cl_2 = 70.90 \, g/mol, and AlCl_3 = 133.33 \, g/mol. 2. Calculate moles of reactants Moles of Al = \frac{34.0}{26.98} = 1.26 \, mol. Moles of Cl_2 = \frac{39.0}{70.90} = 0.55 \, mol. 3. Identify limiting reactant From the reaction, 2Al + 3Cl_2 \to 2AlCl_3, mole ratio is Al : Cl_2 = 2 : 3. Required Cl_2 for 1.26 \, mol \, Al = \frac{3}{2} \times 1.26 = 1.89 \, mol. Available Cl_2 = 0.55 \, mol, so Cl_2 is the limiting reactant. 4. Calculate moles of AlCl_3 produced Mole ratio Cl_2 : AlCl_3 = 3 : 2. Moles of AlCl_3 = \frac{2}{3} \times 0.55 = 0.37 \, mol. 5. Convert moles of AlCl_3 to grams Mass of AlCl_3 = 0.37 \times 133.33 = 49.33 \, g.

Explanation

1. Determine molar masses<br /> Molar mass of $Al = 26.98 \, g/mol$, $Cl_2 = 70.90 \, g/mol$, and $AlCl_3 = 133.33 \, g/mol$.<br /><br />2. Calculate moles of reactants<br /> Moles of $Al = \frac{34.0}{26.98} = 1.26 \, mol$. <br /> Moles of $Cl_2 = \frac{39.0}{70.90} = 0.55 \, mol$.<br /><br />3. Identify limiting reactant<br /> From the reaction, $2Al + 3Cl_2 \to 2AlCl_3$, mole ratio is $Al : Cl_2 = 2 : 3$. <br /> Required $Cl_2$ for $1.26 \, mol \, Al = \frac{3}{2} \times 1.26 = 1.89 \, mol$. <br /> Available $Cl_2 = 0.55 \, mol$, so $Cl_2$ is the limiting reactant.<br /><br />4. Calculate moles of $AlCl_3$ produced<br /> Mole ratio $Cl_2 : AlCl_3 = 3 : 2$. <br /> Moles of $AlCl_3 = \frac{2}{3} \times 0.55 = 0.37 \, mol$.<br /><br />5. Convert moles of $AlCl_3$ to grams<br /> Mass of $AlCl_3 = 0.37 \times 133.33 = 49.33 \, g$.
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