If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the?

Question

If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the balanced equation, how many moles of liquid Br2 remain? Round your answer to 3 significant figures.

2NaI(aq) + Br2(l) ¡æ 2NaBr(aq) + I2(s) If 890 mL of 0.236 M aqueous NaI and 0.560 mol of liquid Br2 are reacted stoichiometrically according to the balanced equation, how many moles of liquid Br2 remain? Round your answer to 3 significant figures.

2NaI(aq) + Br2(l) ¡æ 2NaBr(aq) + I2(s)

Molar Mass (g/mol)
NaI 149.89
Br2 159.81

Density (g/mL)
Br2 3.12

Molar Volume (L)
22.4 at STP

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

I got

Br2 left = 0.56-(0.89x0.236)/2

=.45498

is this right??

Answer 1

Yes but the answer was to be rounded to 3 s.f. Round to 0.455 moles.