calculate the volume of hydrogen liberated at STP when 500cc of 0.5 n sulphuric acid reacts with excess of Zn.
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In order to calculate the volume of hydrogen, Let us analyze the reaction:
Zn + H2SO4-> ZnSO4 + H2
According to the equation, 1 Mole of H2S04 gives one mole of Hydrogen with volume at 22.4 L.
Therefore at 500cc, the total volume of hydrogen = 22.4/4= 5.6L
Zn + H2SO4-> ZnSO4 + H2
According to the equation, 1 Mole of H2S04 gives one mole of Hydrogen with volume at 22.4 L.
Therefore at 500cc, the total volume of hydrogen = 22.4/4= 5.6L
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