Question

Hydrogen gas is produced when magnesium metal reacts with hydrochloric acid. In the lab, a student collects 0.002581 g of hydrogen gas over water at 23°C, and the atmospheric pressure is 794 mmHg. What volume (in mL) of dry hydrogen gas would be collected?

Answer 1

To find the volume of dry hydrogen gas collected, we need to calculate the volume using the ideal gas law equation. First, convert the mass of hydrogen gas collected to moles. Then, plug the values into the ideal gas law equation to find the volume. The volume of dry hydrogen gas collected is 3.28 mL.

Step-by-step explanation:

To find the volume of dry hydrogen gas collected, we need to calculate the volume using the ideal gas law equation: PV = nRT. Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. First, convert the mass of hydrogen gas collected to moles. The molar mass of hydrogen is approximately 2 g/mol. Therefore, the number of moles is 0.002581 g / 2 g/mol = 0.0012905 mol. The temperature conversion from °C to Kelvin: 23°C + 273 = 296 K.

Now, we can plug the values into the ideal gas law equation: (794 mmHg)(V) = (0.0012905 mol)(0.0821 L∙atm/(mol∙K))(296 K). Rearranging the equation to solve for V, we get V = (0.0012905 mol)(0.0821 L∙atm/(mol∙K))(296 K)/(794 mmHg). The unit conversion from mmHg to atm is 1 atm = 760 mmHg. Therefore, V = (0.0012905 mol)(0.0821 L∙atm/(mol∙K))(296 K)/(794 mmHg) * (1 atm/760 mmHg) = 0.00328 L. Finally, convert from liters to milliliters by multiplying by 1000: 0.00328 L * 1000 = 3.28 mL.