Respuesta :
The % of hydrogen peroxide in the solution is 3 %
Explanation
mass percentage is away of representing the concentration of component in a mixture. it is calculated as below;
Mass percent = ( mass of solute/ mass of solution ) x 100
,mass of solute = 0.153 g
mass of solution= 5.02 g
mass percentage is therefore= 0.153 g /5.02 g x 100 = 3%
The percent of hydrogen peroxide [tex]\left({{{\text{H}}_2}{{\text{O}}_2}}\right)[/tex] in the solution is [tex]\boxed{5.75\;\% }[/tex] .
Further Explanation:
The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:
1. Molarity (M)
2. Molality (m)
3. Mole fraction (X)
4. Parts per million (ppm)
5. Mass percent ((w/w) %)
6. Volume percent ((v/v) %)
Mass percent:
It is the ratio of the mass of a component divided by the total mass of the solution, multiplied by 100.
The given reaction is as follows:
[tex]2{{\text{H}}_{\text{2}}}{{\text{O}}_2}\left( l \right) \to 2{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right) + 2{{\text{O}}_2}\left( g \right)[/tex]
The formula to calculate the moles of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] is as follows:
[tex]{\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{O}} = \frac{{{\text{Given mass of }}{{\text{H}}_{\text{2}}}{\text{O}}}}{{{\text{Molar mass of }}{{\text{H}}_{\text{2}}}{\text{O}}}}[/tex] …… (1)
Substitute 0.153 g for the given mass of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] and 18 g/mol for the molar mass of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] in equation (1).
[tex]\begin{aligned}{\text{Moles of }}{{\text{H}}_{\text{2}}}{\text{O}}&=\left({{\text{0}}{\text{.153 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{18 g}}}}}\right)\\&=0.0085{\text{ mol}}\\\end{aligned}[/tex]
According to the stoichiometry of the reaction, two moles of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] reacts to form two moles of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] . So the number of moles of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] becomes equal to that of [tex]{{\text{H}}_{\text{2}}}{\text{O}}[/tex] and therefore the number of moles of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] is 0.0085 mol.
The formula to calculate the mass of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] is as follows:
[tex]{\text{Mass of}}\,{{\text{H}}_{\text{2}}}{{\text{O}}_2}=\left( {{\text{Moles of }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}}\right)\left({{\text{Molar mass of }}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}}\right)[/tex] …… (2)
Substitute 0.0085 mol for the moles of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] and 34 g/mol for the molar mass of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] in equation (2).
[tex]\begin{aligned}{\text{Mass of}}\,{{\text{H}}_{\text{2}}}{{\text{O}}_2}&=\left({{\text{0}}{\text{.0085 mol}}}\right)\left({\frac{{{\text{34 g}}}}{{{\text{1 mol}}}}}\right)\\&=0.28{\text{9 g}}\\\end{aligned}[/tex]
The formula to calculate the mass percent of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] is as follows:
[tex]{\text{Mass}}\;{\text{percent}}=\left({\frac{{{\text{Mass of }}{{\text{H}}_{\text{2}}}{{\text{O}}_2}}}{{{\text{Mass of solution}}}}}\right)\left( {100}\right)[/tex] …… (3)
The mass of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] is 0.289 g.
The mass of the solution is 5.02 g.
Substitute these values in equation (3).
[tex]\begin{aligned}{\text{Mass}}\;{\text{percent}}&=\left({\frac{{{\text{0}}{\text{.289 g}}}}{{{\text{5}}{\text{.02 g}}}}}\right)\left({100}\right)\\&=5.75\;\% \\\end{aligned}[/tex]
Therefore the percent of [tex]{{\text{H}}_{\text{2}}}{{\text{O}}_2}[/tex] is 5.75 %.
Learn more:
1. The mass of ethylene glycol: https://brainly.com/question/4053884
2. Determine the moles of water produced: https://brainly.com/question/1405182
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Concentration terms
Keywords: mass percent, H2O2, O2, H2O, 5.75 %, 5.02 g, mass of solution, mass of H2O2, 0.0085 mol, 34 g/mol, 18 g/mol,
