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10 volume hydrogen peroxide percentage
On the proportion of ten volumes of hydrogen peroxide
Hydrogen peroxide is also a chemical substance. The world is known as "ten volumes of hydrogen peroxide", which refers to the fact that under certain conditions, the volume of oxygen released by its decomposition is ten to one.
When hydrogen peroxide decomposes, according to chemical reasons, the reaction formula can be obtained: $2H_ {2} O_ {2}\ stackrel {catalyst }{=\!=\!=} 2H_ {2} O + O_ {2}\ uparrow $. Therefore, every two molecules of hydrogen peroxide decompose, one molecule of oxygen can be obtained.
Now say ten volumes of hydrogen peroxide, let the volume of hydrogen peroxide solution be $V $, and the volume of oxygen released is $10V $. Under standard conditions, the molar volume of the gas is about $22.4L/mol $. If the volume of oxygen is $10V $, the amount of oxygen substance $n (O_ {2}) =\ frac {10V} {22.4} mol $.
According to the reaction formula, it can be seen that $n (H_ {2} O_ {2}) = 2n (O_ {2}) =\ frac {2\ times10V} {22.4} mol $. The molar mass of hydrogen peroxide $M (H_ {2} O_ {2}) = 34g/mol $, then the mass of hydrogen peroxide $m (H_ {2} O_ {2}) = n (H_ {2} O_ {2}) \ times M (H_ {2} O_ {2}) =\ frac {2\ times10V} {22.4}\ times34g $.
Usually the density of hydrogen peroxide solution is similar to that of water, about $1g/mL $, that is, $1000g/L $, then the mass of hydrogen peroxide solution $m_ {solution} = 1000Vg $.
Therefore, the mass fraction of hydrogen peroxide in a ten-volume hydrogen peroxide solution is $\ omega =\ frac {\ frac {2\ times10V} {22.4}\ times34} {1000V}\ times100\% $, and its proportion can be calculated. This is the proportion of ten volumes of hydrogen peroxide. In practical applications and chemical research, it is clear that this proportion is very important, and it is related to the process and effect of many reactions.
Hydrogen peroxide is also a chemical substance. The world is known as "ten volumes of hydrogen peroxide", which refers to the fact that under certain conditions, the volume of oxygen released by its decomposition is ten to one.
When hydrogen peroxide decomposes, according to chemical reasons, the reaction formula can be obtained: $2H_ {2} O_ {2}\ stackrel {catalyst }{=\!=\!=} 2H_ {2} O + O_ {2}\ uparrow $. Therefore, every two molecules of hydrogen peroxide decompose, one molecule of oxygen can be obtained.
Now say ten volumes of hydrogen peroxide, let the volume of hydrogen peroxide solution be $V $, and the volume of oxygen released is $10V $. Under standard conditions, the molar volume of the gas is about $22.4L/mol $. If the volume of oxygen is $10V $, the amount of oxygen substance $n (O_ {2}) =\ frac {10V} {22.4} mol $.
According to the reaction formula, it can be seen that $n (H_ {2} O_ {2}) = 2n (O_ {2}) =\ frac {2\ times10V} {22.4} mol $. The molar mass of hydrogen peroxide $M (H_ {2} O_ {2}) = 34g/mol $, then the mass of hydrogen peroxide $m (H_ {2} O_ {2}) = n (H_ {2} O_ {2}) \ times M (H_ {2} O_ {2}) =\ frac {2\ times10V} {22.4}\ times34g $.
Usually the density of hydrogen peroxide solution is similar to that of water, about $1g/mL $, that is, $1000g/L $, then the mass of hydrogen peroxide solution $m_ {solution} = 1000Vg $.
Therefore, the mass fraction of hydrogen peroxide in a ten-volume hydrogen peroxide solution is $\ omega =\ frac {\ frac {2\ times10V} {22.4}\ times34} {1000V}\ times100\% $, and its proportion can be calculated. This is the proportion of ten volumes of hydrogen peroxide. In practical applications and chemical research, it is clear that this proportion is very important, and it is related to the process and effect of many reactions.
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