posted by andy .
If 750 mL of 0.893 M aqueous NaBr and 11.6 L of gaseous Cl2 measured at STP are reacted stoichiometrically according to the balanced equation, how many milliliters of 3.93 M aqueous NaCl are produced?
2NaBr(aq) + Cl2(g) ¨ 2NaCl(aq) + Br2(l)
Calculate moles NaBr initially. moles = M x L.
Calculate moles Cl2 initially. Use PV = nRT or the fact that a gas at STP occupies 22.4 L.
Determine the limiting reagent and from that the moles NaCl produced.
Then moles = M x L. You know moles and M, calculate L and convert to mL. Post your work if you get stuck.
Mole of NaBr=(.893M)(.750L)= .6679 mole
mole of Cl2= (1.00atm)(22.4L)=n(.0821)(273.15k) n= .998858 mole
How do u determine the limiting reagent?
2NaBr + Cl2 ==> 2NaCl + Br2
First you transposed the answer for M x L = 0.893 x 0.750 = 0.6697 (not 79).
Next, you didn't calculate moles Cl2 correctly. You COULD have used 11.6/22.4 = 0.518 moles (or you could have used PV = nRT) If used properly that gives you the same answer. The set us is as follows
1atm x 11.6L = n(0.08206)(273) and solve for n.
Use both moles Cl2 and moles NaBr to find moles NaCl produced, then take the smaller number.
0.6697 moles NaBr x (2 moles NaCl/2 moles NaBr) = 0.6697 x 2/2 = 0.6697 moles NaCl produced.
0.518 moles Cl2 x (2 moles NaCl/1 mole Cl2) = 1.036 mols NaCl produced.
The smaller number is the one to use and the NaBr is the limiting reagent; therefore, you can use 0.6697 moles NaCl produced.
Are you to add water to some final volume to prepare 3.93 M NaCl. I ask because all you have present is 0.750 L and that as a volume for 0.6697 moles will not produce a M of 3.93 M.