QuestionJune 5, 2025

Based on the equation how many grams of Br_(2) are required to react completely with 29.2 grams of AlCl_(3) 2AlCl_(3)+3Br_(2)arrow 2AlBr_(3)+3Cl_(2) 48.7 grams 52.6 grams 56.7 grams 61.3 grams

Based on the equation how many grams of Br_(2) are required to react completely with 29.2 grams of AlCl_(3) 2AlCl_(3)+3Br_(2)arrow 2AlBr_(3)+3Cl_(2) 48.7 grams 52.6 grams 56.7 grams 61.3 grams
Based on the equation how many grams of Br_(2) are required to react completely with 29.2 grams of AlCl_(3) 2AlCl_(3)+3Br_(2)arrow 2AlBr_(3)+3Cl_(2) 48.7 grams 52.6 grams 56.7 grams 61.3 grams

Solution
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Answer

61.3 grams Explanation 1. Calculate moles of AlCl_{3} Molar mass of AlCl_{3} is 133.33 \, g/mol. Moles of AlCl_{3} = \frac{29.2}{133.33}. 2. Use stoichiometry to find moles of Br_{2} From the equation, 2 \, mol \, AlCl_{3} reacts with 3 \, mol \, Br_{2}. Therefore, moles of Br_{2} = \frac{3}{2} \times moles of AlCl_{3}. 3. Calculate grams of Br_{2} Molar mass of Br_{2} is 159.808 \, g/mol. Grams of Br_{2} = moles of Br_{2} \times 159.808.

Explanation

1. Calculate moles of $AlCl_{3}$<br /> Molar mass of $AlCl_{3}$ is $133.33 \, g/mol$. Moles of $AlCl_{3} = \frac{29.2}{133.33}$.<br /><br />2. Use stoichiometry to find moles of $Br_{2}$<br /> From the equation, $2 \, mol \, AlCl_{3}$ reacts with $3 \, mol \, Br_{2}$. Therefore, moles of $Br_{2} = \frac{3}{2} \times$ moles of $AlCl_{3}$.<br /><br />3. Calculate grams of $Br_{2}$<br /> Molar mass of $Br_{2}$ is $159.808 \, g/mol$. Grams of $Br_{2} = $ moles of $Br_{2} \times 159.808$.
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