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
see above.
One or two to remmeber, that is.
To solve this problem, we need to use stoichiometry to determine the number of liters of gaseous Cl2 that remain after the reaction.
First, we'll calculate the number of moles of NaBr in the solution:
Molarity of NaBr = 0.283 M
Volume of NaBr = 600 mL = 0.6 L
Number of moles of NaBr = Molarity × Volume
Number of moles of NaBr = 0.283 mol/L × 0.6 L
Number of moles of NaBr = 0.17 moles
According to the balanced equation, the stoichiometric ratio of NaBr to Cl2 is 2:1. This means that 2 moles of NaBr react with 1 mole of Cl2.
Since we have 0.17 moles of NaBr, we need half that amount (0.085 moles) of Cl2 to completely react.
Now, let's calculate the volume of Cl2 required:
Molar volume at STP = 22.4 L/mol
Volume of Cl2 = Number of moles of Cl2 × Molar volume at STP
Volume of Cl2 = 0.085 moles × 22.4 L/mol
Volume of Cl2 = 1.9 L
Therefore, 1.9 liters of gaseous Cl2 were required to completely react with the NaBr solution.
However, we started with 15.9 L of Cl2, so to find the remaining amount, we subtract the volume used:
Remaining volume of Cl2 = Initial volume of Cl2 - Volume of Cl2 used
Remaining volume of Cl2 = 15.9 L - 1.9 L
Remaining volume of Cl2 = 14.0 L
So, there are 14.0 liters of gaseous Cl2 that remain after the reaction.
Please note that when rounding your answer, you should round to three significant figures, so the final answer is 14.0 L.