Respuesta :
Harry started with [tex]\boxed{{\text{5}}{\text{.5 grams}}}[/tex] of [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] hydrated salt.
Further explanation:
Mole is the measure of the amount of substance. Mole is the relation between the mass of the substance and molar mass of a substance. It is defined as the mass of a substance in grams divided by its molar mass (g/mol).
The expression to calculate the number of moles is as follows:
[tex]{\text{Number of moles}}\;=\;\dfrac{{{\text{Given mass}}\left({\text{g}}\right)}}{{{\text{molar mass}}\left({{\text{g/mol}}}\right)}}[/tex]
Therefore, the formula to calculate the mass of the given compound is,
[tex]{\text{Mass}}\left({\text{g}} \right)=\left({{\text{Number of moles}}}\right)\left( {{\text{molar mass}}\left({{\text{g/mol}}}\right)}\right)[/tex]
The molar mass of [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] is 474.3884g/mol.
The molar mass of [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is 258.2050 g/mol.
Therefore, mass of 12[tex]{{\text{H}}_2}{\text{O}}[/tex] molecules is,
[tex]\begin{aligned}{\text{Mass}}\left( {\text{g}} \right)&={\text{Moles}}\times{\text{molar mass of }{{\text{H}}_2}{\text{O}}\\&=12{\text{ mol }}\times 18{\text{ g/mol }}\\&=216{\text{ g}}\\\end{aligned}[/tex]
Since initially in [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] , 1 mole of [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] contained 12 moles of [tex]{{\text{H}}_2}{\text{O}}[/tex] . Thus 258.2050 g of [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] should contains 216 g of [tex]{{\text{H}}_2}{\text{O}}[/tex] before evaporation.
Therefore, the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] contained by 1 g of [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is calculated as,
[tex]{\text{Mass}}\left({\text{g}}\right){\text{ of }}{{\text{H}}_2}{\text{O in 1 g of KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}}{\text{ }}=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left({{\text{S}}{{\text{O}}_{\text{4}}}} \right)}_{\text{2}}}}}[/tex]
Therefore, the mass of [tex]{{\text{H}}_2}{\text{O}}[/tex] contained by 3.0 g of [tex]{\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is calculated as follows:
[tex]\begin{aligned}{\text{Mass}}\left({\text{g}} \right){\text{ of }}{{\text{H}}_2}{\text{O in 3 g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}&=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}\times 3{\text{ g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}\\&=2.50{\text{ g }}\\\end{aligned}[/tex]
This 2.5 g is the mass of water that is evaporated from hydrated salt.
The total mass of hydrated of [tex]{\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}[/tex] taken initially by harry is the sum of the mass of water that is evaporated and mass of dehydrated salt that is left after evaporation process.
[tex]\begin{aligned}{\text{Total mass of hydrated salt}}&={{\text{m}}_{{\text{KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}+{{\text{m}}_{{{\text{H}}_2}{\text{O evaporated}}}}\\&=3.0{\text{ g}}+{\text{2}}{\text{.5 g}}\\&=5.5{\text{ g }}\\\end{aligned}[/tex]
Learn more:
1. Determine the number grams of solute in 500 ml of 0.189 M KOH.: https://brainly.com/question/2847466
2. Rank the gases in decreasing order of effusion:https://brainly.com/question/1946297
Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Mole concept
Keywords: harry, hydrate salt, KAl(SO4)2•12H2O, dehydration process, hydration process, anhydrated salt, KAl(SO4)2, 3.0 gram KAl(SO4)2, grams of start with, 5.5 g.
