Question

. The average concentration of bromine (as bromide) in seawater is \(65 \mathrm{ppm} .\) Calculate (a) the volume of seawater \(\left(d=64.0 \mathrm{lb} / \mathrm{ft}^{3}\right)\) in cubic feet required to produce one kilogram of liquid bromine. (b) the volume of chlorine gas in liters, measured at \(20^{\circ} \mathrm{C}\) and \(762 \mathrm{~mm} \mathrm{Hg}\), required to react with this volume of seawater.

Short answer

Answer: To produce one kilogram of liquid bromine, we need 528.99 ft³ of seawater and 343.60 L of chlorine gas.

Step-by-step solution

Part (a): Finding the volume of seawater needed to produce one kilogram of liquid bromine

First, we'll convert the mass of liquid bromine in kilograms to grams: \(m_{Br_2} = 1\,\text{kg} \times \frac{1000\,\text{g}}{1\,\text{kg}} = 1000\,\text{g}\) We are given the concentration of bromine in seawater as \(65\,\text{ppm}\), which means \(65\,\text{g}\) of bromine is present in \(10^6\,\text{g}\) of seawater: \(\frac{65\,\text{g}\,\text{Bromine}}{10^6\,\text{g}\,\text{Seawater}} = \frac{1000\,\text{g}\,\text{Bromine}}{x\,\text{g}\,\text{Seawater}}\) Now, solve for x: \(x = \frac{1000\,\text{g}\,\text{Bromine} \times 10^6\,\text{g}\,\text{Seawater}}{65\,\text{g}\,\text{Bromine}} = 15384615.38\,\text{g}\,\text{Seawater}\) Convert the mass of seawater to volume using the given density: \(d = \frac{m}{V}\) So, \(V = \frac{m}{d} = \frac{15384615.38\,\text{g}\,\text{Seawater}}{\frac{64.0\,\text{lb}}{1\,\text{ft}^3} \times \frac{453.592\,\text{g}}{1\,\text{lb}}} = 528.99\,\text{ft}^3\)

Part (b): Finding the volume of chlorine gas required to react with the volume of seawater

The balanced chemical reaction between bromide ions (from seawater) and chlorine gas is as follows: \(2\,\text{Br}^{-} + \text{Cl}_2 \rightarrow 2\,\text{Cl}^{-} + \text{Br}_2\) From the stoichiometry of the reaction, we see that \(\text{Cl}_2\) and \(\text{Br}_2\) are in a 1:1 ratio. Therefore, if \(1000\,\text{g}\) of \(\text{Br}_2\) are produced, we need \(1000\,\text{g}\) of \(\text{Cl}_2\). Now we'll use the ideal gas law to find the volume of chlorine gas: \(PV=nRT\) Convert the grams of chlorine gas to moles using the molar mass of chlorine (\(M_{Cl_2} = 70.90\,\text{g/mol}\)): \(n = \frac{1000\,\text{g}}{70.90\,\text{g/mol}} = 14.099\,\text{mol}\) We are given the conditions for chlorine gas: temperature \(T = 20^\circ\,\text{C}\) and pressure \(P = 762\,\text{mmHg}\). We need to convert these units to Kelvin and atmospheres: \(T = 20 + 273.15 = 293.15\,\text{K}\) \(P = 762\,\text{mmHg} \times \frac{1\,\text{atm}}{760\,\text{mmHg}} = 1.0026\,\text{atm}\) Now we can find the volume using the ideal gas law: \(V = \frac{nRT}{P} = \frac{14.099\,\text{mol} \times 0.0821\,\frac{\text{L}\,\text{atm}}{\text{mol}\,\text{K}} \times 293.15\,\text{K}}{1.0026\,\text{atm}} = 343.60\,\text{L}\) In conclusion, to produce one kilogram of liquid bromine, we need \(528.99\,\text{ft}^3\) of seawater and \(343.60\,\text{L}\) of chlorine gas at \(20^\circ\,\text{C}\) and \(762\,\text{mmHg}\).