Multiple Choice 10 points 2Al+3Br_(2)Longrightarrow 2AlBr_(3) How many moles of AlBr3 will be formed from 6.On or Br_(2) 12 mol 0.19 mol 64.8 mol

To determine the number of moles of $AlBr_3$ formed from 6 moles of $Br_2$, we need to use the stoichiometric relationship from the balanced chemical equation:<br /><br />\[ 2Al + 3Br_2 \rightarrow 2AlBr_3 \]<br /><br />From the equation, we see that 3 moles of $Br_2$ produce 2 moles of $AlBr_3$. Therefore, we can set up a proportion to find the number of moles of $AlBr_3$ formed from 6 moles of $Br_2$:<br /><br />\[ \frac{3 \text{ moles of } Br_2}{2 \text{ moles of } AlBr_3} = \frac{6 \text{ of } Br_2}{x \text{ moles of } AlBr_3} \]<br /><br />Solving for \( x \):<br /><br />\[ x = \frac{6 \text{ moles of } Br_2 \times 2 \text{ moles of } AlBr_3}{3 \text{ moles of } Br_2} \]<br /><br />\[ x = 4 \text{ moles of } AlBr_3 \]<br /><br />So, 4 moles of $AlBr_3$ will be formed from 6 moles of $Br_2$.<br /><br />The correct answer is: 4 moles