The mass of sodium chloride that would be produced when 2.3 g of sodium reacts with 0.56 L of chlorine gas at 290K and 1.2 atm. Assume 100% yield.

Answer 1

Find number of moles of chlorine from ideal gas law: n(Cl₂) = PV/RT = (1.2 atm)(0.56 L) / [(0.08206 L atm/mol K)(290K)] = 0.028₂₄ mole Cl₂. Balanced reaction: 2 Na + Cl₂ --> 2 NaCl. Compare yields:

(0.028₂₄ mole Cl₂) * (2 mole NaCl / 1 mole Cl₂) = 0.56₅ mole NaCl from Cl₂.

(2.3g Na) * (1 mole Na /22.99g Na) * (2 mole NaCl / 2 mole Na) = 0.10₀ mole NaCl from Na.

The largest mass of NaCl that can be produced is limited by Cl₂ gas: (0.56₅ mole NaCl) * (58.44g NaCl / 1 mole NaCl) = 33. g NaCl.

Answer 2

Find an answer here

https://youtu.be/2spz4y7iTBs