Ncert Exercises & Solutions

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place

$Z n + 2 H C l \to Z n C l_{2} + H_{2}$

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn=65.3u

Given that, mass of Zn=32.65 g

1 mole of gas occupies=22.7 L volume at STP

Atomic mass of Zn=65.3u

The given equation is

\underset{65.3 g}{Zn} + 2 HCl \to {ZnCl}_{2} + \underset{1 mol = 22.7 L at STP}{H_{2}}

From the above equation, it is clear that

65.3 g Zn, when reacts with HCl, produces

= 22.7 L of H_{2} at STP

∴ 32.65 g Zn, when reacts with HCl, will produce = \frac{22.7 \times 32.65}{65.3} = 11.35 L of H_{2} at STP

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