Question

A reaction of 46.3 g of Na and 36.5 g of Br2 yields 37.9 g of NaBr. What is the percent yield?

2Na(s)+Br₂(g)⟶2NaBr(s)

Answer 1

Answer -

Given,

Mass of Na = 46.3 g

Mass of Br₂ = 36.5 g

Mass of NaBr Yield = 37.9 g

Molar Mass of Na = 23 u

Molar Mass of Br₂ = 159.808 g/mol

Molar Mass of NaBr = 102.894 g/mol

Percentage Yield = ?

2 Na (s) + Br₂ (g) ⟶ 2 NaBr (s) [BALANCED]

We know that,

Moles = Mass / Molar Mass

Moles of Na in 46.3 g = 46.3 g / 23 u = 2.01

Moles of Br₂ in 36.5 g = 36.5 g / 159.808 g/mol = 0.2284 mol

Find Limiting Reagent,

Divide the available moles by the stiochiometric coefficients, the one which is smaller is Limiting Reagent.

For Na = 2.01 mol/2 = 1.005 mol

For Br₂ = 0.2284 mol/1 = 0.2284 mol

So, Br2 is the limiting reagent.

Using Stiochiometry,

It can be analyzed that for 1 mole of Br₂ , 2 moles of NaBr produced.

i.e. moles of NaBr produced = 2 * Moles of Br₂

So, moles of NaBr produced = 2 * 0.2284 mol = 0.4568 mol

also,

Mass = Moles * Molar Mass

Theoretical Mass of NaBr produced = 0.4568 mol * 102.894 g/mol

Theoretical Mass of NaBr produced = 47.0 g

Now,

Percentage Yield = (Actual yield/Theoretical Yield) * 100

Percentage Yield = (37.9 g/47.0 g) * 100

Percentage Yield = 80.64 g [ANSWER]