Question
In the reaction, $$2Al\left( s \right) + 6HCl\left( {aq} \right) \to $$ $$2A{l^{3 + }}\left( {aq} \right) + 6C{l^ - }\left( {aq} \right) + 3{H_2}\left( g \right)$$
A.
$$11.2L{H_2}\left( g \right)$$ $$STP$$ is produced for every mole $$HCl\left( {aq} \right)$$ consumed
B.
$$6\,L\,HCl\left( {aq} \right)$$ is consumed for every $$3\,L\,{H_2}\left( g \right)$$ produced
C.
$$33.6\,L\,{H_2}\left( g \right)$$ is produced regardless of temperature and
pressure for every mole $$Al$$ that reacts
D.
$$67.2\,{H_2}\left( g \right)$$ at $$STP$$ is produced for every mole $$Al$$ that reacts.
Answer :
$$11.2L{H_2}\left( g \right)$$ $$STP$$ is produced for every mole $$HCl\left( {aq} \right)$$ consumed
Solution :
$$2Al\left( s \right) + 6HCl\left( {aq} \right) \to $$ $$2A{l^{3 + }}\left( {aq} \right) + 6C{l^ - }\left( {aq} \right) + 3{H_2}\left( g \right)$$
$$\because \,6\,{\text{moles}}\,{\text{of}}\,HCl\,{\text{produces}}\, = $$ $$3\,{\text{moles}}\,{\text{of}}\,{H_2}$$
$$ = 3 \times 22.4\,L\,{\text{of}}\,{H_2}\,{\text{at}}\,S.T.P$$
$$\therefore 1\,{\text{mole}}\,{\text{of}}\,HCl\,{\text{produces}} = $$ $$\frac{{3 \times 22.4}}{6}L\,\,{\text{of}}\,{H_2}\,{\text{at}}\,{\text{S}}{\text{.T}}{\text{.P}}$$
$$ = 11.2\,L\,\,{\text{of}}\,\,{H_2}\,{\text{at}}\,\,{\text{STP}}$$