Hi, please help me with this question

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


Molar Volume (L)
22.4 at STP

Gas Constant
(L.atm.mol-1.K-1)
0.0821

To solve this question, we need to determine the limiting reactant and then calculate the volume of the remaining gaseous Cl2 measured at STP.

Step 1: Calculate the number of moles of NaBr:
Concentration of NaBr = 0.283 M
Volume of NaBr = 600 mL = 0.6 L (convert mL to L)
Number of moles of NaBr = Concentration × Volume = 0.283 M × 0.6 L = 0.170 moles

Step 2: Calculate the number of moles of Cl2:
Molar mass of Cl2 = 70.906 g/mol
Volume of Cl2 = 15.9 L
Number of moles of Cl2 = Volume ÷ Molar volume = 15.9 L ÷ 22.4 L/mol = 0.71 moles

Step 3: Determine the limiting reactant:
According to the balanced equation, 2 moles of NaBr react with 1 mole of Cl2. Therefore, the mole ratio of NaBr to Cl2 is 2:1.
Since we have 0.170 moles of NaBr and 0.71 moles of Cl2, the moles of Cl2 is in excess.

Step 4: Calculate the number of moles of Cl2 consumed:
From the limiting reactant calculation, we know that 0.170 moles of NaBr react with 0.085 moles of Cl2 (0.170 ÷ 2).

Step 5: Calculate the volume of Cl2 remaining:
Number of moles of Cl2 remaining = Number of moles of Cl2 initially - Number of moles of Cl2 consumed
= 0.71 moles - 0.085 moles
= 0.625 moles

Step 6: Convert the moles of Cl2 remaining to volume at STP:
Volume of Cl2 remaining = Number of moles of Cl2 remaining × Molar volume at STP
= 0.625 moles × 22.4 L/mol
= 13.999 L

Step 7: Round the answer to 3 significant figures:
Volume of Cl2 remaining = 14.0 L

Therefore, the answer is that 14.0 liters of gaseous Cl2 measured at STP remain.

To solve this question, we need to use the balanced equation and the given information about the reactants, NaBr and Cl2.

First, let's calculate the number of moles of NaBr. To do this, we can use the concentration and volume given for NaBr:

Number of moles of NaBr = concentration * volume
= 0.283 M * 0.600 L
= 0.170 moles

According to the balanced equation, 2 moles of NaBr react with 1 mole of Cl2. So, the number of moles of Cl2 required to completely react with NaBr can be obtained by dividing the number of moles of NaBr by 2:

Number of moles of Cl2 required = 0.170 moles / 2
= 0.085 moles

Now, let's convert the number of moles of Cl2 to liters at STP. We can use the molar volume at STP, which is 22.4 L/mol:

Number of liters of Cl2 required = number of moles * molar volume
= 0.085 moles * 22.4 L/mol
= 1.904 L

So, 1.904 L of gaseous Cl2 measured at STP are required to react stoichiometrically with 600 mL of 0.283 M aqueous NaBr.

To find the remaining volume of Cl2 measured at STP, we subtract the required volume from the initial volume of 15.9 L:

Remaining volume of Cl2 = Initial volume - Required volume
= 15.9 L - 1.904 L
= 13.996 L

Rounding to 3 significant figures, the remaining volume of Cl2 measured at STP is 14.0 L.