Calculate the volume of hydrogen liberated at STP. 500 ml, 0.5M sulphuric acid reacts with excess of zinc. According to equation Zn+H2SO4 -Zn(SO4) +H2

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Solution

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2} 1 m o l a r 1 L H_{2} S O_{4} p r o d u c e s 1 m o l e o f h y d r o g e n = 22.4 L 0.5 m o l a r 500 m l (0.5 L) p r o d u c e s = \frac{22.4}{1 \times 1} \times 0.5 \times 0.5 = 5.6 L o f h y d r o g e n a t S T P .$

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Calculate the volume of hydrogen liberated at STP. 500 ml, 0.5M sulphuric acid reacts with excess of zinc. According to equation Zn+H2SO4 -Zn(SO4) +H2

Zn + H2SO4 →ZnSO4 + H21 molar 1 L H2SO4 produces 1 mole of hydrogen = 22.4 L0.5 molar 500 ml (0.5 L) produces = 22.41 × 1 × 0.5 × 0.5 = 5.6 L of hydrogen at STP.

$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2} 1 m o l a r 1 L H_{2} S O_{4} p r o d u c e s 1 m o l e o f h y d r o g e n = 22.4 L 0.5 m o l a r 500 m l (0.5 L) p r o d u c e s = \frac{22.4}{1 \times 1} \times 0.5 \times 0.5 = 5.6 L o f h y d r o g e n a t S T P .$