At S. T. P how many litres of hydrogen can be obtained from the reaction of 500 cm\(^3\) of 0.5 M H\(_2\)SO\(_4\) with excess zinc metal?
(Gram molecular volume of H\(_2\) = 22.4 dm\(^3\))
Zn + H\(_2\)SO\(_4\) → ZnSO\(_4\) + H\(_2\)
1mole : 1mole
n = \(\frac{CV}{1000}\)
= \(\frac{{0.5}\times{500}}{1000}\)
= 0.25 mole : 0.25 mole
n = \(\frac{Volume}{Molar Volume}\)
0.25 = \(\frac{V}{22.4}\)
V = 0.25 x 22.4
V = 5.6 dm\(^3\)
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