Multiple Choice 10 points 2Al+3Br_(2)arrow 2AlBr_(3) How many moles of AlBr_(3) will be formed from 6.0g of Br_(2) 12 mol 0.19 mol 64.8 mol

To determine the number of moles of $AlBr_{3}$ formed from 6.0g of $Br_{2}$, we need to follow these steps:<br /><br />1. Calculate the molar mass of $Br_{2}$.<br />2. Convert the given mass of $Br_{2}$ to moles.<br />3. Use the stoichiometry of the balanced chemical equation to find the moles of $AlBr_{3}$ formed.<br /><br />Step 1: Calculate the molar mass of $Br_{2}$.<br />The molar mass of $Br_{2}$ is the sum of the molar masses of two bromine atoms.<br />Molar mass of $Br_{2}$ = 79.9 g/mol + 79.9 g/mol = 159.8 g/mol<br /><br />Step 2: Convert the given mass of $Br_{2}$ to moles.<br />Given mass of $Br_{2}$ = 6.0 g<br />Moles of $Br_{2}$ = Given mass / Molar mass<br />Moles of $Br_{2}$ = 6.0 g / 159.8 g/mol = 0.0375 mol<br /><br />Step 3: Use the stoichiometry of the balanced chemical equation to find the moles of $AlBr_{3}$ formed.<br />According to the balanced chemical equation, 3 moles of $Br_{2}$ produce 2 moles of $AlBr_{3}$.<br />Moles of $AlBr_{3}$ = (2 moles $AlBr_{3}$ / 3 moles $Br_{2}$) × 0.0375 mol $Br_{2}$<br />Moles of $AlBr_{3}$ = 0.025 mol<br /><br />Therefore, the correct answer is 0.025 mol.