If 390 mL of 0.0410 M aqueous NaI(aq) and 1.36 g of liquid Br2 are reacted stoichiometrically according to the equation, what mass (g) of NaI(aq) remained?

Question

2 NaI(aq) + Br2(l) ¨ 2 NaBr(aq) + I2(s) 
 Molar Mass (g) 
NaI(aq)  149.89  
Br2  159.81

Molar Volume (L)  
22.400 at STP  
Gas Constant 
0.0821

DrBob222 · Answer

1. Convert 390 mL of 0.0410 M NaI to moles NaI. M x L = ?? moles. I found 0.01599 moles NaI.
2. Convert 1.36 g Br2 to moles. moles = g/molar mass. I found 0.008510 moles Br2.
3. Which is the limiting reagent?
a. moles NaBr formed if all NaI is used is
0.01599 x (2 mols NaBr/2 moles NaI) = 0.01599 x 1/1 = 0.01599 moles NaBr.

b. moles NaBr formed if all of the Br2 is used is
0.008510 x (2 moles NaBr/1 mol Br2)= 0.008510 x 2 = 0.01702 moles NaBr.

c. The smaller number is the one to use so NaI is the limiting reagent which means that all of it used and none remains.
Check my numbers. Check my thinking.