How many millilitres of 3 % left( frac{w}{v} | vedclass

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(D) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$.At $STP$,$1 \ mol$ of $O_2$ occupies $22400 \

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$15$

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(D) The decomposition of $H_2O_2$ is given by the reaction: $2H_2O_2(aq) ightarrow 2H_2O(l) + O_2(g)$.At $STP$,$1 \ mol$ of $O_2$ occupies $22400 \ mL$.Therefore,$150 \ mL$ of $O_2$ corresponds to $\frac{150}{22400} \ mol$ of $O_2$.From the stoichiometry,$2 \ mol$ of $H_2O_2$ produces $1 \ mol$ of $O_2$.So,moles of $H_2O_2$ required = $2 	imes \frac{150}{22400} = \frac{300}{22400} \ mol$.Molar mass of $H_2O_2 = 34 \ g/mol$.Mass of $H_2O_2$ required = $\frac{300}{22400} 	imes 34 \approx 0.455 \ g$.$A$ $3 \% \left( \frac{w}{v} ight)$ solution means $3 \ g$ of $H_2O_2$ in $100 \ mL$ of solution.Volume of solution required = $\frac{0.455 \ g}{3 \ g} 	imes 100 \ mL \approx 15.18 \ mL$.Rounding to the nearest integer,the required volume is $15 \ mL$.

How many millilitres of $3 \% \left( \frac{w}{v} \right) H_2O_2$ solution is required to get $150 \ mL$ of oxygen at $STP$?

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