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.
Calculate the volume of hydrogen liberated (Zn=63,1 mole of any gas at STP= 22.4dm3)
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