Question 3.75: The first part of the problem is a simple dilution problem (......
The first part of the problem is a simple dilution problem (M_1\text{V}_1=M_2\text{V}_2). The second part requires the molar mass of the HCl along with the molarity.
\text{a) }M_1=11.7 \ M\quad \text{V}_1=?\quad\quad M_2=3.5 \ M \quad\quad \text{V}_2=5.0\text{ gal} \\ M_1\text{V}_1=M_2\text{V}_2\quad\quad(11.7\text{ M)(V}_1)=(3.5\text{ M)(5.0 gal}) \\ \frac{(3.5\text{ M)(5.0 gal})}{11.7\text{ M}}=\text{V}_1\quad\quad\quad \text{V}_1=1.4957\text{ gallons (unrounded)}
Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 3.0 gallons of water into the container. Add slowly and with mixing 1.5 gallon of 11.7 M HCl into the water. Dilute to 5.0 gallons with water.
\text{b) Volume needed}=(9.55\text{ g HCl})\left\lgroup\frac{1\text{ mol HCl}}{36.46\text{ g HCl}} \right\rgroup \left\lgroup\frac{1\text{ L}}{11.7\text{ mol HCl}} \right\rgroup \left\lgroup\frac{1\text{ mL}}{10^{-3}\text{ L}} \right\rgroup \\ =22.38725=22.4\text{ mL muriatic acid solution}