Formula 1 point When 8.43 liters of chlorine gas is added to aluminum metal how many grams of Aluminum chloride is produced? underline ( )Al +underline ( )Cl_(2) arrow underline ( )AlCl_(3) Type your answer...

## Step 1<br />The balanced chemical equation for the reaction between aluminum and chlorine gas is:<br />### \(2Al + 3Cl_{2} \rightarrow 2AlCl_{3}\)<br />This equation tells us that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride.<br /><br />## Step 2<br />We are given that 8.43 liters of chlorine gas is used. We need to convert this volume to moles using the ideal gas law, which states that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.<br /><br />## Step 3<br />So, the number of moles of chlorine gas is:<br />### \(8.43 \, \text{liters} \times \frac{1 \, \text{mole}}{22.4 \, \text{liters}} = 0.375 \, \text{moles}\)<br /><br />## Step 4<br />From the balanced chemical equation, we know that 3 moles of chlorine gas produce 2 moles of aluminum chloride. Therefore, 0.375 moles of chlorine gas will produce:<br />### \(0.375 \, \text{moles} \times \frac{2 \, \text{moles AlCl}_{3}}{3 \, \text{moles Cl}_{2}} = 0.25 \, \text{moles AlCl}_{3}\)<br /><br />## Step 5<br />The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol. Therefore, the mass of aluminum chloride produced is:<br />### \(0.25 \, \text{moles} \times 133.34 \, \text{g/mol} = 33.34 \, \text{grams}\)