If 0.720 mol of gaseous Cl2 and 390 mL of 0.641 M aqueous NaBr are reacted stoichiometrically according to the balanced equation, how many moles of gaseous Cl2 remain? Round your answer to 3 significant figures.
2NaBr(aq) + Cl2(g) → 2NaCl(aq) + Br2(l)
You know the amount of Cl2. You can find the moles NaBr from M x L = moles. From that you can calculate the amount of NaBr and Cl2 used and from that the amount of either of the reactants in excess. Show your work if you get stuck.
Thanks, i figured it out using the mol ratios of NaBr and Cl2.
Way to go.
To find the number of moles of gaseous Cl2 that remain, we first need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely used up and determines the maximum amount of product that can be formed.
First, let's calculate the number of moles of Br2 that can be formed using the given information:
- From the balanced equation, we know that 2 moles of NaBr react with 1 mole of Cl2 to produce 1 mole of Br2.
- The given amount of NaBr is 390 mL of 0.641 M solution. To find the moles of NaBr, we can use the formula: moles = concentration (M) × volume (L).
- Convert the milliliters to liters: 390 mL ÷ 1000 mL/L = 0.390 L
- Calculate the moles of NaBr: 0.641 M × 0.390 L = 0.25019 moles of NaBr
Since the balanced equation tells us that 2 moles of NaBr react with 1 mole of Cl2 to produce 1 mole of Br2, we can conclude that we need half the amount of Cl2 (0.25019/2 = 0.12509 moles) to react with all the NaBr.
Now, let's compare this value to the initial amount of Cl2:
- The initial amount of Cl2 is given as 0.720 moles.
Since the initial amount of Cl2 (0.720 moles) is greater than the amount needed to react with all the NaBr (0.12509 moles), we can conclude that Cl2 is in excess, and some moles of Cl2 will remain after the reaction.
Finally, to find the number of moles of Cl2 that remain, subtract the amount needed to react from the initial amount:
0.720 moles - 0.12509 moles = 0.59491 moles of Cl2
Rounding to 3 significant figures, the number of moles of gaseous Cl2 that remain is approximately 0.595 moles.