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$H_2O_2$ is marked 22.4 volume.How much of it is required to oxidize 3.5g $H_2S$ gas ?

$(a)\;6\qquad(b)\;7\qquad(c)\;4\qquad(d)\;2$

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Step 1:
$H_2O_2$ is marked 22.4 volume,which means 1litre $H_2O_2$ gives 22.4litre of $O_2$ at NTP.
$2H_2O_2\rightarrow 2H_2O+O_2$
$22400ml$ of $O_2$ is obtained from =68g$H_2O_2$
$22.4ml$ of $O_2$ is obtained from =$\large\frac{68\times 22.4}{22400}$
$\Rightarrow 0.068gH_2O_2$
Step 2:
This amount is present in 1ml $H_2O_2$ solution.
$H_2O_2+H_2S\rightarrow 2H_2O+S$
$34g H_2S$ is oxidized by =68g
$3.5g H_2S$ is oxidized by =$\large\frac{68\times 3.5}{34}$
$\Rightarrow 7g H_2O_2$
Hence (b) is the correct answer.
answered Oct 30, 2013 by sreemathi.v
 

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