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.
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
2 answers
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.
0.62492 mol x 22.4 L/mol = 13.9982 L = 14.0 L Cl2 un-reacted.
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