Problemas
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.
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Rafael
veterano · Tutor durante 10 años
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Respuesta
To solve this problem, we need to use the balanced chemical equation and the given information to calculate the amount of aluminum chloride produced.Given information:- 8.65 liters of chlorine gas (Cl2) is added to aluminum metal (Al).The balanced chemical equation for the reaction is:2Al + 3Cl2 → 2AlCl3Step 1: Calculate the number of moles of chlorine gas (Cl2).Moles of Cl2 = Volume of Cl2 / Molar volume of Cl2Molar volume of Cl2 = 22.4 L/mol (at standard temperature and pressure)Moles of Cl2 = 8.65 L / 22.4 L/mol = 0.385 molesStep 2: Calculate the number of moles of aluminum chloride (AlCl3) produced.According to the balanced equation, 3 moles of Cl2 react with 2 moles of Al to produce 2 moles of AlCl3.Moles of AlCl3 produced = (2/3) × Moles of Cl2Moles of AlCl3 produced = (2/3) × 0.385 moles = 0.257 molesStep 3: Calculate the mass of aluminum chloride (AlCl3) produced.Molar mass of AlCl3 = 133.34 g/molMass of AlCl3 produced = Moles of AlCl3 × Molar mass of AlCl3Mass of AlCl3 produced = 0.257 moles × 133.34 g/mol = 34.2 gramsTherefore, when 8.65 liters of chlorine gas is added to aluminum metal, 34.2 grams of aluminum chloride is produced.