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Cl2 + 2NaBr => Br2 + 2NaCl

One mole Cl2 reacts with 2 moles NaBr

Cl2 = 71

NaBr = 102.9

Molar volume = 22.414 L/mole for ideal gas @STP

3L Cl2 = 3/22.414 = 0.1338 mole

25g NaBr = 25/102.9 = 0.2430 mole

0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction

The NaBr is the limiting reagent

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Q: What reactant is limiting if 3000 cm3 of Cl2 at STP react with a solution containing 25 grams of NaBr?
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