Calculate the volume of hydrogen liberated (Zn=63,1 mole of any gas at STP= 22.4dm3)
To calculate the volume of hydrogen gas liberated, we need to know the stoichiometry of the reaction and the number of moles of the reactant.
Assuming the reaction given is:
Zn + 2HCl -> ZnCl2 + H2
From the stoichiometry of the reaction, we can see that for every 1 mole of Zn, 1 mole of H2 gas is liberated.
Now, if we have 63.1 moles of Zn, we can calculate the volume of H2 gas liberated using the conversion factor provided: 1 mole of any gas at STP = 22.4 dm3.
Therefore, the volume of hydrogen gas liberated will be:
Volume = (Number of moles of H2) x (Conversion factor)
The number of moles of H2 is equal to the number of moles of Zn (based on stoichiometry).
Hence, the volume of hydrogen gas liberated will be:
Volume = 63.1 moles x 22.4 dm3/mole
Volume = 1410.44 dm3
So, the volume of hydrogen gas liberated is 1410.44 dm3.
To calculate the volume of hydrogen gas liberated, we need to know the number of moles of hydrogen gas produced.
Given that 1 mole of Zinc (Zn) liberates 1 mole of hydrogen gas (H2), and the molar mass of Zinc is 63.1 g/mol, we can assume that one mole of Zinc liberates 1 mole of hydrogen gas.
However, the volume of any gas at STP (Standard Temperature and Pressure) is 22.4 L/mol.
Therefore, the volume of hydrogen gas liberated can be calculated as follows:
Volume of Hydrogen Gas (H2) = 22.4 L/mol x 1 mol
Volume of Hydrogen Gas (H2) = 22.4 L
Therefore, the volume of hydrogen gas liberated is 22.4 dm3.