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When the same amount of zinc is treated separately with excess of $H_2SO_4$ and excess of $NaOH$,the ratio of volumes of $H_2$ evolved is

$(a)\; 1 : 1\qquad(b)\;1 : 2\qquad(c)\;2 : 1\qquad(d)\;9 : 4$

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$Zn+H_2SO_4\rightarrow ZnSO_4+H_2$
$Zn+2NaOH\rightarrow Na_2ZnO_2+H_2$
$\therefore $ratio of volumes of $H_2$ evolved is 1 : 1
Hence (a) is the correct answer.
answered Oct 25, 2013 by sreemathi.v

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