How many moles of Chlorine gas are required to produce 5.35 moles of NaCl in this
reaction?

2.675 moles of Chlorine gas are required

Explanation:

Chlorine gas, Cl2, reacts with NaBr to produce NaCl and Br2 as follows:

Cl2 + 2NaBr → 2NaCl + Br2

Where 1 mole of chlorine gas produce 2 moles of NaCl

Thus, to produce 5.35 moles of NaCl are required:

5.35mol NaCl * (1mol Cl2 / 2mol NaCl) =

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

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