It's part of a lab where we reacted magnesium with Hcl to form Hydrogen gas and Magnesium chloride. I have the balanced equation but I have no idea how to find the molar volume of dry Hydrogen gas (without water vapor).
I have the data for the mass of 1.00m of Magnesium Ribbon, the barometer reading of room pressure, room temp(C), vapor pressure of water at room temp, length of Mg used(cm), and the volume of Hydrogen gas (this includes water vapor, I think).
Any help would be appreciated.
The molar mass of hydrogen gas (H2) is 2.0158g. Hydrogen gas is diatomic therefore it's formula is H2. The atomic mass of hydrogen is 1.0079g and multiplying by the two hydrogen atoms is equal to 2.0158g. but i think you want molar volume, which is different
I would first change to liters
assuming (and it's usually true) that pure (or dry) H2 is an ideal gas in this case, then the molar volume of H2 is just the same as the volume of one mol of ideal gas particles. Using the ideal gas equation, the volume is: V = R T / P = (.0821 L atm per K per mol) (room temp) / (room pressure in atm) Remember that the room pressure in atm = ( room pressure in mmHg) / 760 mmHg And be sure to convert the temperature readings from the Celsius scale to the Kelvin scale. if your barometer reads pressures in mmHg, then the reading of the room pressure be more or less equal to 760 mmHg, unless the instrument displays a pressure difference.
Let's do a sample calculation. Pretend the room temperature is about 20 deg C (typical air-conditioned room temp), and the room pressure is 764 mmHg (note that 1 atm is equivalent to 760 mmHg). Then the molar volume of the pure (or dry) H2 gas is: V = R T / P = (.0821 * 293.15 ) / (764 / 760 ) Liter = 23.9 liters Notice that i converted the temperature from 20 deg Celsius to 294.15 Kelvin, and the pressure from mmHg to atm. If you don't know how to do the conversion, refer to any g-chem textbook. The molar volume you get depends on the measurements, but it will all be more or less equal to 24 liters. one mol of ideal gas occupies 24 liters at 1 atm and 25 deg Celsius. this is a handy number about an ideal gas that everyone remembers.
So this is the dry H2 molar volume, but i don't think you need it at all. the reason is that the total pressure of the gas you captured is the sum of the partial pressure of H2 and the partial pressure of water: P (gas mix) = P(H2) + P(H2O) P(gas mix) is the pressure you measure. it should be the same as the room pressure if the water inside the container is level with the water outside the container. So: P(room) = P(H2) + P(H2O) P(H2O) is the partial pressure of the water vapor. because this water vapor is in equilibrium with the water. the water vapor pressure is equal to the vapor pressure of water at the temperature of the lab. So: P(room) = P(H2) + (vapor pressure of H2O at room T) From your lab results, you can find P(H2): P(H2) = P(room) - (vapor pressure of H2O at room T) Then from P(H2), which is the pressure of the dry H2, we use the ideal gas law to find the number of H2 (in mol), like so: number of H2 (in mol) = P(H2) * Volume of container / R / (room temp) What's the volume of the container? It's the same as the volume of the gas mixture because the H2 gas occupies the entire volume (as does the water vapor). Now we have the number of H2, and we can use the balanced chemical reaction to figure out the number of magnesium that reacted to create the H2 that you captured.
i meant the number of magnesium atoms in moles.
i think the balanced equation is: 2 Mg(s) + 2 HCl(aq) -> 2MgCl + H2(g) Not sure if magnesium chloride is very soluble in water or not. By this stoichiometry: number of Mg = number of H2 * (2 mol Mg / 1 mol H2) this is the number of Mg (in moles) that reacted to cause H2 to be created (from HCl).