I got 62.9 mL. Is this correct? I think HCl is the LR.
A piece of solid magnesium is reacted with hydrochloric acid to form hydrogen gas:
Mg (s) + 2HCl (aq) ---> MgCl2 (aq) + H2 (g)
What volume of hydrogen gas (in mL) is collected over water at 25 degrees Celsius by reaction of 0.450 g of Mg (AW= 24.3 g/mol) with 5.00 mL of 1.0 M HCl? The barometer records an atmospheric pressure of 758 torr and the vapor pressure of water at this temperature is 23.35 torr.
Close but not there. Since that other problem was close, too, perhaps you are not rounding right. Post your work here and I will check it. MY answer is 63.2 mL.
424.25×10^3 pa
To find the volume of hydrogen gas collected over water at 25 degrees Celsius, we need to use the ideal gas law:
PV = nRT
Where:
P = total pressure (atmospheric pressure + vapor pressure)
V = volume of the gas
n = number of moles of the gas (we can find this using stoichiometry)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25 degrees Celsius = 298 K)
First, we need to calculate the number of moles of hydrogen gas produced by the reaction between magnesium and hydrochloric acid. We can use the balanced equation to determine the stoichiometry of the reaction:
1 mol of Mg reacts with 2 mol of HCl to produce 1 mol of H2
The molar mass of magnesium (Mg) is 24.3 g/mol, so:
0.450 g of Mg * (1 mol Mg / 24.3 g Mg) = 0.0185 mol of Mg
Since the reaction stoichiometry is 1:1 between Mg and H2, we have 0.0185 mol of H2.
Now, let's calculate the total pressure in the system. The total pressure is the sum of the atmospheric pressure and the vapor pressure of water:
Total pressure = atmospheric pressure + vapor pressure
Total pressure = 758 torr + 23.35 torr
Total pressure = 781.35 torr
Next, let's substitute these values into the ideal gas law equation:
PV = nRT
V * (781.35 torr) = (0.0185 mol H2) * (0.0821 L·atm/(mol·K)) * (298 K)
Now, we can solve for V:
V = (0.0185 mol H2) * (0.0821 L·atm/(mol·K)) * (298 K) / (781.35 torr)
V = 0.444 L
Finally, let's convert the volume from liters to milliliters:
V = 0.444 L * (1000 mL / 1 L)
V = 444 mL
Therefore, the volume of hydrogen gas collected over water at 25 degrees Celsius is 444 mL.