Question

aqueous hydrobromic acid (hbr) will react with solid sodium hydroxide (naoh) to produce aqueous sodium bromide (nabr) and liquid water (h₂o). suppose 61. g of hydrobromic acid is mixed with 58.2 g of sodium hydroxide. calculate the maximum mass of sodium bromide that could be produced by the chemical reaction. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

$$\ce{HBr + NaOH -> NaBr + H2O}$$

Step2: Calculate the molar masses

• Molar mass of $\ce{HBr}$: $M_{HBr}=1 + 79.904=80.904\space g/mol$
• Molar mass of $\ce{NaOH}$: $M_{NaOH}=22.99 + 16+ 1.008 = 39.998\space g/mol$
• Molar mass of $\ce{NaBr}$: $M_{NaBr}=22.99+79.904 = 102.894\space g/mol$

Step3: Calculate the number of moles of reactants

• Moles of $\ce{HBr}$: $n_{HBr}=\frac{m_{HBr}}{M_{HBr}}=\frac{61\space g}{80.904\space g/mol}\approx0.754\space mol$
• Moles of $\ce{NaOH}$: $n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}=\frac{58.2\space g}{39.998\space g/mol}\approx1.455\space mol$

Step4: Determine the limiting reactant

From the balanced equation, the mole ratio of $\ce{HBr}$ to $\ce{NaOH}$ is $1:1$. Since $n_{HBr}

Step5: Calculate the moles of $\ce{NaBr}$ produced

From the balanced equation, $n_{NaBr}=n_{HBr} = 0.754\space mol$

Step6: Calculate the mass of $\ce{NaBr}$ produced

$m_{NaBr}=n_{NaBr}\times M_{NaBr}=0.754\space mol\times102.894\space g/mol\approx77.6\space g$

Answer:

$77.6\space g$