If 600 mL of 0.283 M aqueous NaBr and 15.9 L of gaseous Cl2 measured at STP are reacted stoichiometrically according to the balanced equation, how many liters of gaseous Cl2 measured at STP remain? Round your answer to 3 significant figures.
2NaBr(aq) + Cl2(g) ¡æ 2NaCl(aq) + Br2(l)
Molar Mass (g/mol)
NaBr 102.89
Cl2 70.906
Density (g/mL)
-
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
is this rigt??
moles of Cl2 = 15.9 / 22.4 = 0.710
excess Cl2 = .710 - .0849 = 0.625 moles or 0.625 x 22.5 liter = 14.1 liters at STP
You have worked the problem correctly but I would not round until the final step. In addition, the last 22.5 you used should be 22.4.
mols NaBr = 0.600 x .283 = 0.1698 (I've carried it to one more place than is allowed; then I will round at the end.).
moles Cl2 needed = 0.1698/2 = 0.08490.
moles Cl2 = 15.9/22.4 = 0.70982.
0.70982-0.08490 = 0.62492 moles Cl2 un-reacted.
0.62492 mol x 22.4 L/mol = 13.9982 grams = 14.0 grams Cl2 un-reacted.
Obviously, the unit is liters and not grams..
0.62492 mol x 22.4 L/mol = 13.9982 L = 14.0 L Cl2 un-reacted.
To solve this problem, we can use the concept of stoichiometry to determine the amount of excess Cl2 remaining after the reaction.
First, let's calculate the moles of Cl2 present in the 15.9 L of gaseous Cl2 measured at STP. Given that the molar volume at STP is 22.4 L/mol, we divide the volume by the molar volume:
moles of Cl2 = 15.9 L / 22.4 L/mol = 0.710 mol
Now we need to determine the limiting reactant. To do this, we can compare the moles of Cl2 to the moles of NaBr.
According to the balanced equation, the stoichiometric ratio between Cl2 and NaBr is 1:2. This means that 1 mole of Cl2 reacts with 2 moles of NaBr.
The moles of NaBr can be calculated using the molarity (0.283 M) and volume (600 mL = 0.600 L):
moles of NaBr = molarity x volume
moles of NaBr = 0.283 M x 0.600 L = 0.170 mol
Since the stoichiometric ratio is 1:2, the Cl2 is the limiting reactant since there are fewer moles of Cl2 (0.710 mol) compared to NaBr (0.170 mol).
To find the excess Cl2 remaining, we subtract the moles of NaBr that reacted from the total moles of Cl2:
excess Cl2 = moles of Cl2 - moles of NaBr
excess Cl2 = 0.710 mol - 0.170 mol = 0.540 mol
Finally, to convert the excess moles of Cl2 to liters at STP, we can use the molar volume at STP:
excess liters of Cl2 = excess moles of Cl2 x molar volume at STP
excess liters of Cl2 = 0.540 mol x 22.4 L/mol = 12.096 L
Rounding to 3 significant figures, the answer is 12.1 liters of gaseous Cl2 measured at STP remain after the reaction.
So, the correct answer is 12.1 liters of gaseous Cl2 remain at STP.