Question

aqueous hydrobromic acid (hbr) will react with solid sodium hydroxide (naoh) to produce aqueous sodium bromide (nabr) and liquid water (h₂o). suppose 51. g of hydrobromic acid is mixed with 7.81 g of sodium hydroxide. calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

$HBr(aq)+NaOH(s)
ightarrow NaBr(aq) + H_2O(l)$

Step2: Calculate the molar masses

The molar mass of $HBr$ is $M_{HBr}=1.01 + 79.90=80.91\ g/mol$. The molar mass of $NaOH$ is $M_{NaOH}=22.99+16.00 + 1.01 = 40.00\ g/mol$.

Step3: Calculate the moles of $NaOH$

$n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}=\frac{7.81\ g}{40.00\ g/mol}=0.19525\ mol$

Step4: Determine the moles of $HBr$ that react with $NaOH$

From the balanced - equation, the mole - ratio of $HBr$ to $NaOH$ is $1:1$. So, the moles of $HBr$ that react with $0.19525\ mol$ of $NaOH$ is $n_{HBr\ reacted}=0.19525\ mol$.

Step5: Calculate the mass of $HBr$ that reacts

$m_{HBr\ reacted}=n_{HBr\ reacted}\times M_{HBr}=0.19525\ mol\times80.91\ g/mol = 15.8077775\ g$

Step6: Calculate the mass of $HBr$ left over

$m_{HBr\ left}=m_{HBr\ initial}-m_{HBr\ reacted}=51\ g - 15.8077775\ g=35.1922225\ g\approx35\ g$ (rounded to two significant digits)

Answer:

$35\ g$