A tank contains a mixture of \(52.5 \mathrm{~g}\) oxygen gas and \(65.1 \mathrm{~g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\). The total pressure in the tank is \(9.21\) atm. Calculate the partial pressures of each gas in the container.

We are given the temperature as 27°C. In order to use the Ideal Gas Law, we need to convert this to Kelvin. This conversion is done by adding 273.15 to the Celsius temperature: \( T(K) = T(^\circ C) + 273.15 \) \( T(K) = 27 + 273.15 = 300.15 K \)

In order to find the number of moles for each gas, we use the given masses and their molar masses. The molar mass of oxygen (O₂) is 32.00 g/mol and the molar mass of carbon dioxide (CO₂) is 44.01 g/mol: For oxygen gas: \( n_{O_2} = \frac{m_{O_2}}{M_{O_2}} \) \( n_{O_2} = \frac{52.5 g}{32.00 \mathrm{~g/mol}} \) \( n_{O_2} = 1.6406 \mathrm{~mol} \) For carbon dioxide gas: \( n_{CO_2} = \frac{m_{CO_2}}{M_{CO_2}} \) \( n_{CO_2} = \frac{65.1 g}{44.01 \mathrm{~g/mol}} \) \( n_{CO_2} = 1.4790 \mathrm{~mol} \)

Dalton's Law states that the total pressure of a mixture of gases is the sum of their partial pressures, which gives us: \( P_{total} = P_{O_2} + P_{CO_2} \) We know the total pressure is 9.21 atm, and we need to find the partial pressures of each gas.

Using the Ideal Gas Law for each gas, we have: For oxygen gas: \( P_{O_2} = \frac{n_{O_2}RT}{V} \) For carbon dioxide gas: \( P_{CO_2} = \frac{n_{CO_2}RT}{V} \) Since both gases share the same volume and temperature in the container, we can rearrange and solve for the partial pressures of each gas: \( P_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{CO_2}} \times P_{total} \) \( P_{O_2} = \frac{1.6406}{1.6406 + 1.4790} \times 9.21 ~\mathrm{atm} \) \( P_{O_2} = 4.94 ~\mathrm{atm} \) Similarly, for carbon dioxide: \( P_{CO_2} = \frac{n_{CO_2}}{n_{O_2} + n_{CO_2}} \times P_{total} \) \( P_{CO_2} = \frac{1.4790}{1.6406 + 1.4790} \times 9.21 ~\mathrm{atm} \) \( P_{CO_2} = 4.27 ~\mathrm{atm} \) Thus, the partial pressures of oxygen and carbon dioxide in the tank are 4.94 atm and 4.27 atm, respectively.