Formula 1 point When 8.65 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.

To solve this problem, we need to use the balanced chemical equation and the given information to calculate the amount of aluminum chloride produced.<br /><br />Given information:<br />- 8.65 liters of chlorine gas (Cl2) is added to aluminum metal (Al).<br /><br />The balanced chemical equation for the reaction is:<br />2Al + 3Cl2 → 2AlCl3<br /><br />Step 1: Calculate the number of moles of chlorine gas (Cl2).<br />Moles of Cl2 = Volume of Cl2 / Molar volume of Cl2<br />Molar volume of Cl2 = 22.4 L/mol (at standard temperature and pressure)<br />Moles of Cl2 = 8.65 L / 22.4 L/mol = 0.385 moles<br /><br />Step 2: Calculate the number of moles of aluminum chloride (AlCl3) produced.<br />According to the balanced equation, 3 moles of Cl2 react with 2 moles of Al to produce 2 moles of AlCl3.<br />Moles of AlCl3 produced = (2/3) × Moles of Cl2<br />Moles of AlCl3 produced = (2/3) × 0.385 moles = 0.257 moles<br /><br />Step 3: Calculate the mass of aluminum chloride (AlCl3) produced.<br />Molar mass of AlCl3 = 133.34 g/mol<br />Mass of AlCl3 produced = Moles of AlCl3 × Molar mass of AlCl3<br />Mass of AlCl3 produced = 0.257 moles × 133.34 g/mol = 34.2 grams<br /><br />Therefore, when 8.65 liters of chlorine gas is added to aluminum metal, 34.2 grams of aluminum chloride is produced.