Question

Aqueous hydrobromic acid HBr

 reacts with solid sodium hydroxide NaOH

 to produce aqueous sodium bromide NaBr

 and liquid water H

2

O

. If 11.2

g

of sodium bromide is produced from the reaction of 26.7

g

 of hydrobromic acid and 17.6

g

 of sodium hydroxide, calculate the percent yield of sodium bromide. 

Round your answer to 3

 significant figures

Answer 1

HBr + NaOH → NaBr + H₂O

M(HBr) = 80.91 g/mol

$n(HBr) = \frac{26.7}{80.91} = 0.329 \;mol$

M(NaOH) = 39.99 g/mol

$n(NaOH) = \frac{17.6}{39.99} = 0.440 \;mol$

n(HBr) < n(NaOH)

So, HBr is a limiting reactant.

n(NaBr) = n(HBr) = 0.329 mol

m(NaBr) = 102.89 g/mol

$m(NaBr) = 0.329 \times 102.89 = 33.85 \;g$

Proportion:

33.85 – 100 %

11.2 – x

$x = \frac{11.2 \times 100}{33.85} = 33.0 \; \%$

Answer: 33.0 %