Mg + 2HCl -> MgCl2 + H2
Over an interval of 1.00s, the mass of Mg changes by -0.011g
What is the corresponding rate of consumption of HCl (in mol/s)?
I got 9.0E-4 mol/s and that's the right answer. But I'm having problems with the next question.
Calculate the corresponding rate of production of H2 in L/s at 20°C and 101 kPa. Tip: 1.00 mol of any gas occupies a volume of 24.0 L at 20°C and 101 kPa.
What I have so far:
delta[H2]/(delta)t =
i.e. I have no idea what to do here...
for each mole of Mg consumed, the rate of H2 is the same. That will be in moles/sec
to change that to volume, use
V=n*24 where n is in moles/sec, so volume will be in liters/sec
To calculate the rate of production of H2 in L/s, you need to determine the moles of H2 produced per second.
From the balanced equation, you know that 1 mol of Mg produces 1 mol of H2. Therefore, the rate of H2 production is equal to the rate of Mg consumption.
In the given question, it is mentioned that the mass of Mg changes by -0.011 g over 1.00 s.
To determine the moles of Mg consumed, you need to use the molar mass of Mg, which is 24.31 g/mol.
Moles of Mg consumed = (Change in mass of Mg) / (Molar mass of Mg)
= (-0.011 g) / (24.31 g/mol)
Now, you have the moles of Mg consumed per second.
To convert this to moles of H2 produced per second, you use the stoichiometry of the balanced equation.
Moles of H2 produced per second = Moles of Mg consumed per second
Finally, to convert the moles of H2 produced per second to liters per second, you use the ideal gas law equation:
PV = nRT
Here, P = 101 kPa, V = volume in liters, n = moles of H2 produced, R = 8.314 L·kPa/mol·K (gas constant), and T = 20°C = 293.15 K (temperature in Kelvin).
Rearranging the equation, you get:
V = (nRT) / P
Substituting the known values, you can calculate the volume of H2 produced per second, which will give you the rate of production of H2 in L/s.
To calculate the corresponding rate of production of H2 in L/s, we'll need to use the stoichiometry of the balanced equation and the given information.
First, we know that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of MgCl2 and 1 mole of H2.
We are given the rate of change in mass of Mg (-0.011 g) over a time interval of 1.00 s. Keep in mind that the mass of Mg also represents the amount (in moles) of Mg consumed.
So, we can convert the change in mass of Mg to moles using its molar mass. The molar mass of Mg is 24.31 g/mol.
Change in moles of Mg = -0.011 g / 24.31 g/mol = -4.52 x 10^-4 mol
Since the stoichiometry tells us that the ratio of moles of HCl consumed to moles of H2 produced is 2:1, we can calculate the corresponding change in moles of H2.
Change in moles of H2 = (2 mol H2 / 1 mol Mg) * (-4.52 x 10^-4 mol) = -9.04 x 10^-4 mol
Now, we need to convert the change in moles of H2 to volume in liters. We can use 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 in Kelvin.
Given:
- Pressure (P) = 101 kPa (which is equivalent to 101000 Pa)
- Temperature (T) = 20°C = 20 + 273.15 = 293.15 K
- 1 mole of any gas occupies a volume of 24.0 L at 20°C and 101 kPa (as stated in the tip)
Now we can calculate the volume of H2 produced using the ideal gas law:
V = (n * R * T) / P
V = (-9.04 x 10^-4 mol * 8.314 L·kPa·mol^-1·K^-1 * 293.15 K) / 101000 Pa
V = -2.08 x 10^-5 L
Note: The negative sign indicates a decrease in volume due to the consumption of H2.
Finally, we can determine the rate of production of H2 in L/s by dividing the change in volume by the time interval of 1.00 s:
Rate of production of H2 = (-2.08 x 10^-5 L) / (1.00 s)
Therefore, the corresponding rate of production of H2 is approximately -2.08 x 10^-5 L/s.