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)
To answer this question, we need to use the stoichiometry of the balanced equation and the given information to determine the number of milliliters of 3.93 M aqueous NaCl that are produced.
Step 1: Calculate the number of moles of NaBr.
Using the given concentration and volume of the aqueous solution of NaBr:
750 mL * (0.893 mol/L) = 669.75 mol NaBr
Step 2: Use the stoichiometry of the balanced equation to determine the number of moles of Cl2 required.
From the balanced equation, we can see that 2 moles of NaBr react with 1 mole of Cl2 to form 2 moles of NaCl. Therefore, the stoichiometry ratio is 2:1.
Since we have 669.75 moles of NaBr, we will need half that amount of Cl2:
669.75 mol NaBr / 2 = 334.875 mol Cl2
Step 3: Convert the amount of Cl2 in moles to liters at STP.
To convert moles to liters at STP, we can use the ideal gas law: PV = nRT.
At STP, the values for P (pressure) and T (temperature) are 1 atm and 273 K, and the value for R (ideal gas constant) is 0.0821 L*atm/(mol*K).
Using these values, we can solve for the volume (V):
V = nRT / P
V = (334.875 mol)(0.0821 L*atm/(mol*K))(273 K) / 1 atm
V = 7695.3 L
Step 4: Calculate the number of moles of NaCl produced.
Since the reaction has a stoichiometric ratio of 2:2 for NaCl and NaBr, the number of moles of NaCl produced will be the same as the number of moles of NaBr used. Thus, the number of moles of NaCl produced is also 669.75 mol.
Step 5: Calculate the volume of the 3.93 M aqueous NaCl solution.
To find the volume, we need to use the concentration and the number of moles of NaCl produced:
Volume = number of moles / concentration
Volume = 669.75 mol / (3.93 mol/L) = 170.29 L
Step 6: Convert the volume of NaCl to milliliters.
Finally, we can convert the volume from liters to milliliters:
170.29 L * 1000 mL/L = 170290 mL
Therefore, approximately 170,290 milliliters of the 3.93 M aqueous NaCl solution are produced.
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.
Using NaBr:
0.6697 moles NaBr x (2 moles NaCl/2 moles NaBr) = 0.6697 x 2/2 = 0.6697 moles NaCl produced.
Using Cl2:
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.