Chemistry, 08.07.2019 14:30 christinapim

Suppose harry begins with the hydrate kal(so4)2·12h2o. after dehydration he finds that he is left with 3.0 g of the an-hydrate kal(so4)2. how many grams did he start with?

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Suppose harry begins with the hydrate kal(so4)2·12h2o. after dehydration he finds that he is left wi

Answer from: asauerha

Harry started with $\boxed{{\text{5}}{\text{.5 grams}}}$ of ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ 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:

${\text{Number of moles}}\;=\;\dfrac{{{\text{Given mass}}\left({\text{g}}\right)}}{{{\text{molar mass}}\left({{\text{g/mol}}}\right)}}$

Therefore, the formula to calculate the mass of the given compound is,

${\text{Mass}}\left({\text{g}} \right)=\left({{\text{Number of moles}}}\right)\left( {{\text{molar mass}}\left({{\text{g/mol}}}\right)}\right)$

The molar mass of ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ is 474.3884g/mol.

The molar mass of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is 258.2050 g/mol.

Therefore, mass of 12 ${{\text{H}}_2}{\text{O}}$ molecules is,

$\begin{aligned}{\text{Mass}}\left( {\text{g}} \right)&={\text{Moles}}\times{\text{molar mass of }{{\text{H}}_2}{\text{O}}\\&=12{\text{ mol }}\times 18{\text{ g/mol }}\\&=216{\text{ g}}\\\end{aligned}$

Since initially in ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ , 1 mole of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ contained 12 moles of ${{\text{H}}_2}{\text{O}}$ . Thus 258.2050 g of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ should contains 216 g of ${{\text{H}}_2}{\text{O}}$ before evaporation.

Therefore, the mass of ${{\text{H}}_2}{\text{O}}$ contained by 1 g of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is calculated as,

${\text{Mass}}\left({\text{g}}\right){\text{ of }}{{\text{H}}_2}{\text{O in 1 g of KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}}{\text{ }}=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left({{\text{S}}{{\text{O}}_{\text{4}}}} \right)}_{\text{2}}}}}$

Therefore, the mass of ${{\text{H}}_2}{\text{O}}$ contained by 3.0 g of ${\text{KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is calculated as follows:

$\begin{aligned}{\text{Mass}}\left({\text{g}} \right){\text{ of }}{{\text{H}}_2}{\text{O in 3 g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}&=\dfrac{{{\text{216 g of }}{{\text{H}}_2}{\text{O}}}}{{258.2050{\text{ g of KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}\times 3{\text{ g of KAl}}{\left({{\text{S}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}\\&=2.50{\text{ g }}\\\end{aligned}$

This 2.5 g is the mass of water that is evaporated from hydrated salt.

The total mass of hydrated of ${\text{KAl}}{\left( {{\text{S}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}} \cdot{\text{12}}{{\text{H}}_{\text{2}}}{\text{O}}$ 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.

$\begin{aligned}{\text{Total mass of hydrated salt}}&={{\text{m}}_{{\text{KAl}}{{\left( {{\text{S}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}+{{\text{m}}_{{{\text{H}}_2}{\text{O evaporated}}}}\\&=3.0{\text{ g}}+{\text{2}}{\text{.5 g}}\\&=5.5{\text{ g }}\\\end{aligned}$

Learn more:

1. Determine the number grams of solute in 500 ml of 0.189 M KOH.:

2. Rank the gases in decreasing order of effusion:

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.

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Suppose harry begins with the hydrate kal(so4)2·12h2o. after dehydration he finds that he is left wi...

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