Q. In the reaction, $2Al(s) + 6HCl(aq) \rightarrow 2{Al}^{3 +}(aq) + 6{Cl}^{-}(aq) + 3H_{2}(g)$
A) 6 L HCl ($aq$) is consumed for every $3\ L\ H_{2}(g)$ produced
B) $33.6\ L\ H_{2}(g)$ is produced regardless of temperature and pressure for every mole Al that reacts
C) $67.2\ L\ H_{2}(g)$ at STP is produced for every mole Al that reacts
D) $11.2\ L\ H_{2}(g)$ at STP is produced for every mole HCl($aq$) consumed
Explanation
(d)
$2Al(s) + 6HCl(aq) \rightarrow 2{Al}^{3 +}(aq) + 6{Cl}^{-}(aq) + 3H_{2}(g)$
$3 \times 22.4\ L\ H_{2}(g)$ at STP is produced by 6 moles of HCl ($aq$)
Hence, 11.2 L $H_{2}\ (g)$ at STP is produced by 1 mole HCl ($aq$).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.