Question

aqueous hydrochloric acid (hcl) reacts with solid sodium hydroxide (naoh) to produce aqueous sodium chloride (nacl) and liquid water (h₂o). if 1.90 g of sodium chloride is produced from the reaction of 3.3 g of hydrochloric acid and 6.3 g of sodium hydroxide, calculate the percent yield of sodium chloride. be sure your answer has the correct number of significant digits in it.

Explanation:

Step1: Write the balanced chemical equation

$HCl(aq)+NaOH(s)
ightarrow NaCl(aq) + H_2O(l)$

Step2: Calculate the molar masses

The molar mass of $HCl$ is $M_{HCl}=1.01 + 35.45=36.46\ g/mol$. The molar mass of $NaOH$ is $M_{NaOH}=22.99+16.00 + 1.01 = 40.00\ g/mol$. The molar mass of $NaCl$ is $M_{NaCl}=22.99+35.45 = 58.44\ g/mol$

Step3: Determine the limiting reactant

The number of moles of $HCl$, $n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{3.3\ g}{36.46\ g/mol}\approx0.0905\ mol$. The number of moles of $NaOH$, $n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}=\frac{6.3\ g}{40.00\ g/mol}= 0.1575\ mol$. From the balanced - equation, the mole ratio of $HCl$ to $NaOH$ is $1:1$. Since $n_{HCl}

Step4: Calculate the theoretical yield of $NaCl$

From the balanced equation, the mole ratio of $HCl$ to $NaCl$ is $1:1$. So the number of moles of $NaCl$ produced theoretically, $n_{NaCl - theoretical}=n_{HCl}=0.0905\ mol$. The theoretical mass of $NaCl$, $m_{NaCl - theoretical}=n_{NaCl - theoretical}\times M_{NaCl}=0.0905\ mol\times58.44\ g/mol\approx5.29\ g$

Step5: Calculate the percent yield

The percent yield formula is $\text{Percent Yield}=\frac{m_{NaCl - actual}}{m_{NaCl - theoretical}}\times100\%$. Given $m_{NaCl - actual}=1.90\ g$. So $\text{Percent Yield}=\frac{1.90\ g}{5.29\ g}\times100\%\approx36.0\%$

Answer:

$36.0\%$