Question

Determine the percent yield forthe reaction between 45.9 g ofNaBr and excess chlorine gas toproduce 12.8 g of NaCl and anunknown quantity of bromine gas.

Answer 1

Step 1:

The reaction must be written and balanced:

2 NaBr (aq) + Cl2 (g) → 2 NaCl (aq) + Br2 (g)

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Step 2:

The limiting reactant here is NaBr. The excess agent is the Cl2 gas

The limiting reactant is already specified.

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Step 3:

The actual yield is provided = 12.8 g NaCl

The theoretical yield must be calculated

2 NaBr (aq) + Cl2 (g) → 2 NaCl (aq) + Br2 (g)

The molar mass NaBr) 103 g/mol

The molar mass NaCl) 58.4 g/mol

Procedure:

2 x 103 g NaBr ------ 2 x 58.4 g NaCl

45.9 g NaBr ------ X = 26.0 g NaCl = theoretical yield

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Step 4:

% yield = actual yield/theoretical yield x 100 = (12.8 g/26.0 g) x 100 = 49.2 % approx

Answer: % yield = 49.2 %