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If 73.5 g of liquid Br2 and 330 mL of 1.18 M aqueous NaI are reacted stoichiometrically according to the balanced equation, how many grams of liquid Br2 remain? Round your answer to 3 significant figures.

2NaI(aq) + Br2(l) → 2NaBr(aq) + I2(s)

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    mols NaI = 0.330 x 1.18 = 0.389
    How much Br2 will it use? Look at the coefficients in the balanced equation. It will need half of that; therefore, 0.389/2 = ? mols Br2 use. How many grams is that? g = moles Br2 used x molar mass.
    75.3 g initially - g used = g remaining. Check all of my numbers. I rounded here and there.

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