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.

Question

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

Answer 1

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

Answer 2

see above.

Answer 3

One or two to remmeber, that is.