Calculate the volume of H2 produced at 300 K and 98.65 kPa when 48 g of Mg
reacts with excess HCl according to the following equation:
Mg(s) �2HCl(aq) £ MgCl(aq) �H2(g)
This problem is worked with the same procedure as the last one (with Na).
To calculate the volume of H2 produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in Pascals (Pa)
V = volume in cubic meters (m^3)
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (K)
First, let's calculate the number of moles of H2 produced. We can use the balanced chemical equation to determine the mole ratio between Mg and H2:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)
From the equation, we can see that one mole of Mg reacts with one mole of H2. Therefore, the number of moles of H2 produced will be equal to the number of moles of Mg.
To calculate the number of moles, we need to know the molar mass of Mg. The molar mass of Mg is 24.31 g/mol. Thus:
Number of moles = mass (g) / molar mass (g/mol)
Number of moles of Mg = 48 g / 24.31 g/mol = 1.975 mol
Since the reaction is carried out with excess HCl, all the Mg reacts, resulting in the formation of the same number of moles of H2.
Now, let's calculate the volume of H2 using the ideal gas law equation. We have T = 300 K, P = 98.65 kPa = 98.65 * 10^3 Pa, and n = 1.975 mol:
V = nRT / P
V = (1.975 mol) * (8.314 J/(mol·K)) * (300 K) / (98.65 * 10^3 Pa)
Now, we need to convert the volume from cubic meters to liters by multiplying by 1000:
V = V * 1000
Finally, we can calculate the volume of H2 produced:
V = (1.975 mol) * (8.314 J/(mol·K)) * (300 K) / (98.65 * 10^3 Pa) * 1000
After performing the calculation, we get the volume of H2 produced in liters.