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Choose the Best Answer · Q19

Q.During the decomposition of H2O2H_2O_2 to give dioxygen, 48 g O2O_2 is formed per minute at a certain point of time. The rate of formation of water at this point is

(a) 0.75 mol min−10.75\ \text{mol min}^{-1}
(b) 1.5 mol min−11.5\ \text{mol min}^{-1}
(c) 2.25 mol min−12.25\ \text{mol min}^{-1}
(d) 3.0 mol min−13.0\ \text{mol min}^{-1}
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Step 1. The decomposition is 2H2O2(l)→2H2O(l)+O2(g)2H_2O_2(l)\rightarrow2H_2O(l)+O_2(g), so Rate=12d[H2O]dt=d[O2]dt\text{Rate}=\dfrac{1}{2}\dfrac{d[H_2O]}{dt}=\dfrac{d[O_2]}{dt}, i.e. d[H2O]dt=2d[O2]dt\dfrac{d[H_2O]}{dt}=2\dfrac{d[O_2]}{dt}. …

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