Answer to Question #230849 in General Chemistry for Bob

Question

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H_2O). Suppose 9.71 g of hydrobromic acid is mixed with 6.5 g of sodium hydroxide. Calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. Be sure your answer has three significant digits.

Answer 1

HBr + NaOH → NaBr + H₂O

m(HBr) = 9.71 g

M(HBr) = 80.91 g/mol

$n= \frac{m}{M}$

n(HBr) $= \frac{9.71}{80.91}=0.120 \;mol$

n(NaOH) = 6.5 g

M(NaOH) = 39.99 g/mol

n(NaOH) $= \frac{6.5}{39.99}= 0.162 \;mol$

According to the reation equation we need one mole of NaOH per one mole of Hbr, but in this case we have only 0.120 mol of HBr and 0.162 mol of NaOH. So, HBr is a limiting reactant.

Also according to the reaction equation

n(NaBr) = n(HBr) = 0.120 mol

M(NaBr) = 102.89 g/mol

m(NaBr) $= 0.120 \times 102.89 = 12.34 \;g$

Answer: 12.3 g