A) \[0.34\]and \[0.68\]
B) \[0.32\]and \[0.36\]
C) \[0.34\] and \[0.36\]
D) \[0.32\]and \[0.68\]
Correct Answer: B
Solution :
[b] \[\frac{-d[{{H}_{2}}{{O}_{2}}]}{2dt}=\frac{d[{{H}_{2}}{{O}_{2}}]}{2dt}=\frac{d[{{O}_{2}}]}{dt}\] [a] \[\frac{d[{{O}_{2}}]}{dt}=\frac{1}{2}\times \frac{-dt[{{H}_{2}}{{O}_{2}}]}{dt}\times \frac{Mw\,\,of\,\,{{O}_{2}}}{Mw\,\,of\,\,{{H}_{2}}{{O}_{2}}}\] \[=\frac{1}{2}\times 0.68kg\,h{{r}^{-1}}\times \frac{32\times {{10}^{-3}}kg}{34\times {{10}^{-3}}kg}\] \[=0.32\,kg\,h{{r}^{-1}}\] [b] \[\frac{d[{{H}_{2}}O]}{dt}=\frac{2}{2}\times \frac{-d[{{H}_{2}}{{O}_{2}}]}{dt}\times \frac{Mw\,\,of\,\,{{H}_{2}}O}{Mw\,\,of\,\,{{H}_{2}}{{O}_{2}}}\] \[=0.68\,kg\,h{{r}^{-1}}\times \frac{18\times {{10}^{-3}}kg}{34\times {{10}^{-3}}kg}\] \[=0.36\,kg\,h{{r}^{-1}}\]
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