Question

aqueous hydrobromic acid (hbr) reacts with solid sodium hydroxide (naoh) to produce aqueous sodium bromide (nabr) and liquid water (h₂o). if 1.38 g of sodium bromide is produced from the reaction of 4.0 g of hydrobromic acid and 0.68 g of sodium hydroxide, calculate the percent yield of sodium bromide. be sure your answer has the correct number of significant digits in it.

Explanation:

Step1: Write the balanced chemical equation

$$\ce{HBr(aq) + NaOH(s) -> NaBr(aq) + H2O(l)}$$

Step2: Calculate the molar masses

• Molar mass of $\ce{NaOH}$: $M_{NaOH}=22.99 + 16.00+ 1.01=40.00\ \text{g/mol}$
• Molar mass of $\ce{NaBr}$: $M_{NaBr}=22.99 + 79.90 = 102.89\ \text{g/mol}$
• Molar mass of $\ce{HBr}$: $M_{HBr}=1.01+79.90 = 80.91\ \text{g/mol}$

Step3: Determine the limiting reactant

• Moles of $\ce{NaOH}$: $n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}=\frac{0.68\ \text{g}}{40.00\ \text{g/mol}} = 0.017\ \text{mol}$
• Moles of $\ce{HBr}$: $n_{HBr}=\frac{m_{HBr}}{M_{HBr}}=\frac{4.0\ \text{g}}{80.91\ \text{g/mol}}\approx0.0494\ \text{mol}$

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

Step4: Calculate the theoretical yield of $\ce{NaBr}$

From the balanced equation, mole ratio of $\ce{NaOH}$ to $\ce{NaBr}$ is $1:1$. So moles of $\ce{NaBr}$ produced (theoretical) $n_{NaBr}^{theo}=n_{NaOH} = 0.017\ \text{mol}$
Theoretical mass of $\ce{NaBr}$: $m_{NaBr}^{theo}=n_{NaBr}^{theo}\times M_{NaBr}=0.017\ \text{mol}\times102.89\ \text{g/mol}\approx1.75\ \text{g}$

Step5: Calculate the percent yield

Percent yield formula: $\text{Percent Yield}=\frac{m_{NaBr}^{actual}}{m_{NaBr}^{theo}}\times100\%$
Given $m_{NaBr}^{actual} = 1.38\ \text{g}$ and $m_{NaBr}^{theo}\approx1.75\ \text{g}$
$\text{Percent Yield}=\frac{1.38\ \text{g}}{1.75\ \text{g}}\times100\%\approx78.9\%$

Answer:

$79\%$