Answer:
Explanation:
I'm assume that 3 Br2 + 6 OH¹-5 Br¹ + BrO3¹ + 3 H₂O is meant to read:
3 Br2 + 6 OH^-1 = 5 Br^-¹ + BrO3^-¹ + 3 H₂O
The balanced equation tells us that 3 moles of Br2 will react with 6 moles of OH^-1 to produce 5 moles of Br^-1.
The first step is to determine whether the Br2 or OH^-1 are limiting reagents. That is, is there enough of each to complete the reaction, with none left over.
We need 3 moles of Br2 for every 6 moles of OH^-1, a molar ratio of 1/2 (Br2/OH).
We are given 4.68 moles of Br2 and 8.12 moles of OH^-1. That is a ratio of 4.68/8.12 or 0.5764. This is higher than the ratio of 1/2 or 0.5 that is required. That means we have more than enough Br2. The limiting reagent is the OH^-1. Once it is consumed, the reaction stops and we are left with some unreacted Br2.
So we need the molar ratio of the OH^-1 to the Br from the balanced equation: We see that 6 moles of OH^-1 are required to produce 5 moles of Br, a 6/5 molar ratio.
Therefore, we may assume all 8.12 moles of the limiting reagent, OH^-1, will be consumed to produce *6/5) that amount of Br.
(8.12 moles OH^-1)*((6 moles Br)/(5 moles OH^-1)) = 9.75 moles of Br.
For curiosity's sake, we can determine the amount of unreacted Br2. 8.12 moles of OH^-1 would require (8.12 moles OH)*(1/2) = 4.06 moles of Br2.
4.68 moles starting Br2
4.06 moles consumed
0.62 moles remaining Br2
