This is a limiting reagent problem.
Determine moles Br2. You have that in the problem. Determine moles NaI from moles = M x L. Look at the equation. It will taake twice as much NaI to react so if you have twice as much NaI as Br2, then Br2 will be the limiting reagent and you will have NaI remain. From the way the problem is stated, I assume NaI is the limiting reagent and Br2 will remain. You can tell from the moles of each. Post your work if you get stuck.
If 0.670mol of liquid Br2 and 460mL of 1.72M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many moles of liquid Br2 remain? Round answer to 3 sig figs.
2NaI(aq) + Br2(l) -> 2NaBr(aq) + I2(s)
Additional Information:
MM of Br2 = 159.81 g/mol
MM of NaI = 149.89 g/mol
Density of Br2 = 3.12g/mL
Molar Volume = 22.4L at STP
Gas Constant = 0.0821
1 answer
1 answer