25.0 grams of chlorine reacts with 29.0 grams of sodium 2Na+Cl2→2NaCl find the Percent Yield, Theoretical Yield, Limiting Reactant, Excess Reactant, and Actual Yield.

Answer 1

Most of the problem can be answered. Compare moles of NaCl possibly produced from given masses:

(25.0 g Cl₂) * (1 mole Cl₂ /70.90 g Cl₂) * (2 mole NaCl / 1 mole Cl₂) = 0.705₂ mole NaCl.

(29.0 g Na) * (1 mole Na / 22.99 g Na) * (2 mole NaCl / 2 mole Na) = 1.26₁ mol NaCl.

Chlorine is Limiting Reactant because its amount is not sufficient to react all the sodium.

The Excess Reactant Na = (1.26₁ - 0.705 = 0.556 mole) * (22.99 g Na /1 mole Na) = 12.8 g Na.

Theoretical Yield of NaCl, from consuming all chlorine: (0.705₂ mole NaCl) * (58.44 g NaCl) = 41.2 g NaCl.

The Actual Yield is a laboratory measurement, not derivable from any of the above. If available, then Percent Yield = 100 * actual / theoretical.