If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the balanced equation, how many moles of liquid Br2 remain? Round your answer to 3 significant figures.
2NaI(aq) + Br2(l) ¡æ 2NaBr(aq) + I2(s)
Molar Mass (g/mol)
NaI 149.89
Br2 159.81
Br2 3.12
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
To determine how many moles of liquid Br2 remain after the reaction, we need to consider the stoichiometry of the balanced equation and the initial amounts of the reactants.
From the balanced equation, we can see that the stoichiometric ratio between NaI and Br2 is 2:1. This means that for every 2 moles of NaI, 1 mole of Br2 is required for the reaction.
First, let's calculate the number of moles of NaI present in the solution using the given volume (890 mL) and molarity (0.236 M):
Moles of NaI = Volume (L) x Molarity
= 0.890 L x 0.236 mol/L
≈ 0.210 mol
Since the stoichiometric ratio between NaI and Br2 is 2:1, we can determine the number of moles of Br2 required for the reaction by dividing the number of moles of NaI by 2:
Moles of Br2 required = Moles of NaI / 2
= 0.210 mol / 2
= 0.105 mol
The amount of moles of Br2 given (0.560 mol) is greater than the moles of Br2 required (0.105 mol), so we can conclude that the given amount of Br2 is in excess.
To find out how many moles of Br2 remain, we subtract the moles of Br2 required from the given amount of moles:
Moles of Br2 remaining = Moles of Br2 given - Moles of Br2 required
= 0.560 mol - 0.105 mol
= 0.455 mol
Therefore, the number of moles of liquid Br2 that remain after the reaction is approximately 0.455 mol.
Note: Remember to round your answer to 3 significant figures, so the result is 0.455 mol.