Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place: Zn +2 HCl arrow ZnCl 2+ H 2 What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with HCl ?

Question

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place: Zn +2 HCl arrow ZnCl 2+ H 2 What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with HCl ? (A) 10.03 L (B) 11.35 L (C) 11.57 L (D) 9.53 L

Tardigrade Admin · Accepted Answer

underset text65.3 gZn+2HCl → ZnCl2+ underset text22.7 L H2 According to the equation, 65.3 g of Zn liberates = 22.7 L of H2 32.65 g of Zn will liberate = (22.7/65.3) ×32.65 = 11.35 L

Q. Hydrogen gas is prepared in the laboratory by reacting dilute $H Cl$ with granulated zinc. Following reaction takes place:
$Z n + 2 H Cl \to Z n C l_{2} + H_{2}$
What would be the volume of hydrogen gas is liberated at STP when $32.65 g$ of zinc reacts with $H Cl$ ?

$10.03 L$

$11.35 L$

$11.57 L$

$9.53 L$

$65.3 g Z n + 2 H Cl \to Z n C l_{2} + 22.7 L H_{2}$
According to the equation, $65.3 g$ of $Z n$ liberates $= 22.7 L$ of $H_{2}$
$32.65 g$ of $Z n$ will liberate $= \frac{22.7}{65.3} \times 32.65 = 11.35 L$

Q. Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place: Zn+2HCl→ZnCl2​+H2​ What would be the volume of hydrogen gas is liberated at STP when 32.65g of zinc reacts with HCl ?

10.03L

11.35L

11.57L

9.53L