Answer to Question #120987 in Chemistry for Aisha Sayed

Question

Sodium iodide and chlorine gas undergo a single displacement reaction, as represented by the following equation: 2Nal (s) +Cl 2(g) I 2(s) +2NaCl (s) If 4.0 g of sodium iodide and 4.0 g of chlorine gas are added to the reaction chamber, what is the actual yield of sodium chloride if the percentage yield of the reaction is 67%

Answer 1

Solution:

The balanced chemical equation:

2Nal(s) + Cl₂(g) = I₂(s) + 2NaCl(s)

According to the chemical equation: n(NaI)/2 = n(Cl₂) = n(NaCl)/2

1) Calculate moles:

Moles of NaI = Mass of NaI / Molar mass of NaI

The molar mass of NaI is 149.89 g/mol.

Moles of NaI = (4.0 g) / (149.89 g/mol) = 0.0267 mol

Moles of Cl₂ = Mass of Cl₂ / Molar mass of Cl₂

The molar mass of Cl₂ is 70.906 g/mol.

Moles of Cl₂ = (4.0 g) / (70.906 g/mol) = 0.0564 mol

2) Divide by coefficients of balanced equation:

Sodium iodide (NaI) ⇒ (0.0267 mol) / (2 mol) = 0.01335 mol

Chlorine gas (Cl₂) ⇒ (0.0564 mol) / (1 mol) = 0.0564 mol

Sodium iodide (NaI) is the lower value. It is the limiting reagent.

3) Use NaI : NaCl molar ratio:

1 is to 1 as 0.0267 mol is to X

X = 0.0267 mol of NaCl produced

4) Determine grams of NaCl:

Mass of NaCl = Moles of NaCl × Molar mass of NaCl

The molar mass of NaCl is 58.44 g/mol.

Mass of NaCl = (0.0267 mol) × (58.44 g/mol) = 1.56035 g = 1.56 g

5) Actual yield of NaCl = Mass of NaCl × The percentage yield of the reaction

Actual yield of NaCl = (1.56 g) × (0.67) = 1.0452 g = 1.05 g

Answer: 1.05 g is the actual yield of sodium chloride (NaCl).