A) One mol of \[{{H}_{2}}\]
B) 1/2 mol of \[{{H}_{2}}\]
C) 2/3 mol of \[{{O}_{2}}\]
D) both 1/2 mol of\[{{H}_{2}}\] and 1/2 mol of \[{{O}_{2}}\]
Correct Answer: B
Solution :
[b] \[\underset{\begin{smallmatrix} 1\,mole \\ \frac{1}{2}mole \end{smallmatrix}}{\mathop{Mg}}\,+2HCl\to MgC{{l}_{2}}+\underset{\begin{smallmatrix} \,\,\,1\,mole \\ \,\,\,\frac{1}{2}mole(12g\,of\,Mg=\frac{1}{2}mol) \end{smallmatrix}}{\mathop{{{H}_{2}}}}\,\uparrow \]
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