Question

pre - laboratory exercise: 1. define the following terms: amphoteric a substance that can act both as an acid and base hydrate a compound that has water chemically bonded to it 2. a student reacts 1.005 g al with 50 ml of 4m naoh and an excess of h₂so₄ in solution to produce 15.20 g of sodium alum. a. calculate the molar mass of sodium alum, naal(so₄)₂·12h₂o. b. what is the theoretical yield of sodium alum? (the molar ratio between al and sodium alum is 1 to 1.) c. what is the percent yield of sodium alum?

Explanation:

Step1: Calculate molar mass of sodium alum

The molar - mass of each element: $M_{Na}=22.99\ g/mol$, $M_{Al}=26.98\ g/mol$, $M_{S}=32.07\ g/mol$, $M_{O}=16.00\ g/mol$, $M_{H}=1.01\ g/mol$.
For $NaAl(SO_{4})_{2}\cdot12H_{2}O$:
\[

$$\begin{align*} M&=M_{Na}+M_{Al}+2M_{S}+(2\times4 + 12)M_{O}+24M_{H}\\ &=22.99+26.98 + 2\times32.07+(8 + 12)\times16.00+24\times1.01\\ &=22.99+26.98+64.14 + 320+24.24\\ &=474.35\ g/mol \end{align*}$$

\]

Step2: Calculate moles of Al

The molar mass of Al is $M_{Al}=26.98\ g/mol$. The moles of Al, $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{1.005\ g}{26.98\ g/mol}=0.03725\ mol$.
Since the molar ratio between Al and sodium alum is 1:1, the moles of sodium alum produced theoretically, $n_{theo}=n_{Al}=0.03725\ mol$.

Step3: Calculate theoretical yield of sodium alum

The theoretical yield, $m_{theo}=n_{theo}\times M$, where $M = 474.35\ g/mol$. So $m_{theo}=0.03725\ mol\times474.35\ g/mol = 17.67\ g$.

Step4: Calculate percent yield of sodium alum

The percent yield is given by the formula $\text{Percent Yield}=\frac{m_{actual}}{m_{theo}}\times100\%$. Given $m_{actual}=15.20\ g$ and $m_{theo}=17.67\ g$.
$\text{Percent Yield}=\frac{15.20\ g}{17.67\ g}\times100\% = 86.02\%$

Answer:

a. The molar mass of sodium alum, $NaAl(SO_{4})_{2}\cdot12H_{2}O$ is $474.35\ g/mol$.
b. The theoretical yield of sodium alum is $17.67\ g$.
c. The percent yield of sodium alum is $86.02\%$.